Math 411 gloss, fall 2008

Monday, December 8, 2008

In chapter 6, problem 2 was done (at great length) by Mr. Kowalick (actually, by Mr. Leven, as his surrogate). A standard way to "check" that a continuous function is always 0 is to integrate its square and see what the result is.
In chapter 6, problem 3 was done by Mr. Kiria.
In chapter 6, problem 5 was done by acclamation, mostly, and then cleaned up by Mr. Leven.
In chapter 6, problem 10 a), b), c) was done by Mr. Baldwin. This is a famous result.
In chapter 6, problem 11 was done by Ms. Zhang.
In chapter 6, problem 15 was done by Mr. Thorp. Indeed this is a mathematical form of the Heisenberg Uncertainty Principle.
In chapter 6, problem 16 concerning the Riemann ζ function was rapidly discussed by Ms. Slusky. More than you want to know, but only an introduction, about ζ is here.

Thursday, December 4, 2008

We proved the result that a continuous function of a Riemann integrable function is Riemann integrable. The proof given was copied directly from the text, but the lecturer attempted to give some insight. The text's A and B were called good and bad intervals. On the good intervals, the function doesn't wiggle too much. On the bad intervals, it could wiggle a great deal, but then the total length of the bad intervals is rather small.

Then some corollaries were given (products, absolute values). We discussed the Fundamental Theorem of Calculus.

An attempt was made to discuss curves. Somehow this got sidetracked as the instructor told students about space-filling curves. This is a continuous mapping of, say, the unit interval in R1 onto the unit square in R2. Here is a Wikipedia article. Such curves were first constructed by Giuseppe Peano, and a nice example was invented by I.J. Schoenberg.

The following results tell you a bit more about what the space-filling curves do not (!).

• If f=(f1,f2) is a differentiable function (so the two components are differentiable) then f([0,1]) is "small" in the following sense: given ε>0, there is a final collection of rectangles in R2 which contain f([0,1]) so that the sum of the areas of these rectangles is less than ε. (This is sometimes referred to by the phrase: "f([0,1]) has content 0".) The proof of this just uses the Mean Value Theorem.

Therefore the space-filling curves must not be differentiable (at least in large parts of the domain). The coordinate functions must resemble, for example, the Takagi function mentioned earlier in class: very "strange".

• If f is 1-1, then the interior (in R2!) of f([0,1]) is ∅. Unlike the previous remark, which really does follow easily from the Mean Value Theorem, I don't know any "easy" way to prove this. The follows from a standard result of topology called invariance of domain, which is usually verified with tools from algebraic topology.

Tuesday, November 25, 2008

The plan for the last 3 meetings of the course (today included) is the following:

Discussion of some of these results in the textbook:
     6.11, 6.12, 6.13, 6.15, 6.16, 6.17.

Finish this discussion, and discuss 6.19, 6.20, and 6.21 (and maybe 6.27).

Next Monday, the last day of class
Students would analyze the solutions of problems 2, 3, 5, 10 (a,b,c), 11, 15, 16, preparing for the final exam.

We did indeed discuss some parts of these results:
6.12, 6.15, 6.16, and a version of 6.17. I proved a version of 6.17 with a weaker hypothesis. Also I indicated why the derivative of the Heaviside function (a jump) could be/should be the Dirac delta function. The "generalized Riemann integral" I mentioned towards the end of class is frequently called the Henstock-Kurzweil integral.

Tuesday, November 25, 2008

Why study Riemann-Stieltjes integration?
A student may validly wonder what the heck is to be gotten out of looking at this variant of the integral? Everything "looks" sort of the same, but ... Here are some reasons.
  1. This makes one reconsider the original definitions and theorems, and maybe check them more carefully. So even the "old" stuff you may be confident about is checked carefully.
  2. These integrals are used in probability and in mathematical finance. They are natural there.
  3. R-S integrals allow a more unified treatment of integrals and sums. Everyone wonders at the parallel logical structure of integration and summation. To a great extent, this is verified through use of the R-S formulations.
  4. Most significantly, the idea of a positive measure is a very general part of how to integrate, very abstract and powerful. It turns out that every positive measure corresponds to a Riemann-Stieltjes integral. This is certainly not entirely obvious (and for sure not obvious if you don't know what a positive measure is). But the R-S integral is a way of realizing and computing with an abstract concept.

Basic definitions
As in the text: f is a bounded function on [a,b]; α is an increasing bounded function on that interval; P is a partition of [a,b]; Δαj which is how big α thinks the subinterval is; U(P,f,α) and L(P,f,α) are the upper and lower sums; abf dα and abf dα are the upper and lower Riemann-Stieltjes integrals with respect to α.
In order to keep from going crazy in html, I will refer to the upper R-S integral as U∫abf dα and to the lower R-S integral as L∫abf dα. sigh...

A bounded function f on [a,b] is Riemann-Stieltjes integrable with respect to α (brief written as f∈R(α)) if U∫abf dα=L∫abf dα. The common value if it exists is called the Riemann-Stieltjes integral of f with respect to α.

We showed that if f is continuous on [a,b], then f∈R(x). R(x) is also called just R in the text. These are the usual Riemann integrable functions. The verification used uniform continuity. I remarked (and this came up again later) that actually the verification was valid for any eligible α, that is, any α which is increasing in [a,b]. So any continuous f is in R(α). Necessary and sufficient conditions for a bounded f to be in R were given by Lebesgue and were mentioned in the last lecture and a link was given there to a careful statement and proof.

The Heaviside function
We took α to be defined piecewise by α(x)=0 if x<0 and α(x)=1 if x>0. We investigated what Δαj was, and which functions were in R(α) for the interval [-1,1]. We discovered a much simpler necessary and sufficient condition: f∈R(α) if and only if f is left-continuous at 0. This means limx→0-f(x) exists and equals f(0). The proof, once the definitions have been waded through, is not that difficult.

Adding a point to a partition
We returned to the prosaic general theory by verifying that adding a point to a partition increases (not strictly increases, necessarily!) the lower sums and decreases the upper sums. Then we learned that every lower sum is less than or equal to every upper sum. And we learned that the lower L-S integral is always less than or equal to the upper L-S integral.

Criteria for L-S integrability
We considered the statement: given ε>0, there is a partition Pε so that U(Pε,f,α)-L(Pε,f,α)<ε. This turns out to be equivalent to f∈R(α). We also considered Riemann sums for the Riemann-Stieltjes integral. The sums we have looked at, defined with sups and infs in the subintervals of partitions, are sometimes called Darboux sums. But it turns out that Riemann sums, which rely on f(sample point) instead, also work in more or less the same fashion. I discussed this, and sort of proved it. A detailed proof is in the textbook.

Test taking, etc.
The instructor offered some comments.

Thursday, November 20, 2008

The instructor continued trudging through the text, briefly mentioning the derivative of vector-valued functions. The definition implies that f:[a,b]→Rn is differentiable if and only if (Note: f(x)=(f1,f2,...,fn) is an n-tuple of "scalar" functions) each of the components fj is differentiable.

We saw the counterexample to the strongest form of the Mean Value Theorem (eix with x in [0,2π]). So a version of MVT with equality is false. You see, R1 is very nice. For topology folks, it is the only Rn where the order topology and the Euclidean metric topology are the same, so there are many coincidences. Also a version of L'Hop fails.

However, the MVT is really used in analysis as an inequality. In R1, the result is |f´(x)-f´(y)|≤(Constant)|x-y| where f is differentiable between x and y, and "Constant" is some upper bound on |f´(ξ)| for ξ between x and y. We use this Lipschitz estimate, as we previously called it, in lots of ways. Now the text has a marvelous an efficient proof of Theorem 5.19 giving what turns out to be a best possible (although not so stated) such estimate for f:[a,b]→Rn when f is differentiable. Well, that proof looks magical but it comes from functional analysis and is an example of a big machine. Let me give a less efficient but non-magical proof.

Proof of a MVT estimate for differentiable f:[a,b]→Rn
Here f=(f1,f2,...,fn) is an n-tuple of differentiable scalar (real-valued) functions. We want to estimate |f(x)-f(y)|=|(f1(x),f2(x),...,fn(x))-(f1(y),f2(y),...,fn(y))| and we will use a definitely low-tech trick that, for example, is common in multivariable calculus: change one variable or coordinate at a time. So define
Please realize that g0(x,y)=f(x) and gn(x,y)=f(y). I may have not defined this precisely (hah!) correctly in class -- I am sorry. Almost nothing is precise in class.

Then write |f(x)-f(y)|=|j=0n-1gj+1(x,)-gj(x,y)|. Notice that each of the differences of this telescoping sum has a change in one variable. Use the triangle inequality to overestimage this sum by the sum of the magnitudes of the differences. Now let's analyze each of the magnitudes of the differences.

So |gj+1(x,)-gj(x,y)|=|fj+1(x)-fj+1(y)| because the Euclidean norm or metric takes the square root of the sum of the squares of the differences, and all of the other coordinates have difference equal to 0. But by the "ordinary" MVT, |fj+1(x)-fj(y)|≤Cj+1|x-y| where Cj+1 is some overestimate of |fj+1´(&xi)| when ξ is between x and y. Therefore |f(x)-f(y)|≤|&sumj=1nCj||x-y|. So indeed we have a Lipschitz estimate: |f(x)-f(y)|≤Constant|x-y|. The Constant here is some L1 overestimate of the size of f´ between x and y. The text shows that there is some ξ between x and y so that |f´(ξ)| can be taken as the Constant (the same ξ in each coordinate, and the L2 size is what's there. That is more efficient. Please take a look.

Presentation by Ms. Hood
Ms. Hood discussed the solution of several problems about differentiation. Her work was prepared with the help of Mr. Baldwin. Here is a very nice exposition which she prepared.

Magic, magic, magic ...
The result presented can be given a very general form, and is used in many circumstances. The generalization is known with varous names. It can be called the Contraction Mapping Theorem or the Banach Fixed Point Theorem. Stefan Banach was was once of the early originators and expositors of functional analysis. So what is the result?

Contraction mapping theorem
Suppose X is a complete metric space, and f:X→X is a contraction. That is, there is a constant K with 0≤K<1 so that d(x,y)≤Kd(x,y) for all points x and y in X. Then f has a unique fixed point, p: f(p)=p, and, if x is any point in X, the sequence {xj} defined by x1=x and xj+1=f(xj) for j≥1 always converges with limit equal to p.

Rather than a proof here, I refer you to the proof in the Wikipedia article. The proof is extravagantly simple, and, in fact, the whole setup is so simple (X and f and K) that the theorem has enormous uses in virtually all areas of mathematics.

Please read the proof.

For example, it can be used to provide very neat existence and uniqueness results in differential equations. It even has applications in algebra.

If we "relax" the hypotheses to d(f(x),f(y))<d(x,y) then there need not be a fixed point as Ms. Hood shows. However, if X is compact, then this slightly weaker hypothesis does imply the result. You could prove this! (Look at the continuous real-valued function x→d(x,f(x)) on the compact metric space. What is its [achieved!] minimum? Why is the minimizing point unique?)

Completeness is needed. For example, x→(1/2)x is a contraction on (0,1) but it has no fixed point. Frequently compact metric spaces are used in applications, and this is o.k. since compact metric spaces are always complete.

Moving on
We finally began to move on to integration. I sketched where we were going. I indicated (I'll be more formal later) the ideas of Riemann which are now familiar to us from calculus and which can be restated with only very slight modification in a totally rigorous way.

Suppose f is a bounded real-valued functon on [a,b]. Then we defined partition and upper and lower sums of f with respect to the partition. And the sup of the lower sums over all choices of (which must exist because of the word "bounded"!) is called the lower integral, while the inf of the upper sums is called the upper integral.

Now there's a trick which shows that any upper sum is greater than or equal to any lower sum (I'll go through this later, but it basically involves taking a "common refinement" of the two (possibly distinct) partitions involve and then observing that "adding" one point to a partition (weakly) increases lower sums and (weakly) decreases upper sums. Therefore the lower integral is always less than or equal to the upper integral. When these integrals are equal, we declare that f on [a,b] is Riemann integrable and the common value is called ∫abf.

Which functions are Riemann integrable?
Well, let's consider some examples.

  1. Suppose f is continuous on [a,b]. Then f is Riemann integrable on [a,b].
    This follows easily by using uniform continuity. We need a result approximating f by nice piecewise constant functions.
  2. Are there any integrable functions which are not continuous? (Remember our functions here are always bounded -- we are trying to understand the proper Riemann integral). Sure enough, students responded with an example like this: define on, say, [0,1], the function f(x)=0 if x<(1/2) and f(x)=1 otherwise. Then if we take a partition like [0,(1/2)-ε,(1/2)+ε,1], the difference between the upper and lower sums will be 2ε for any positive ε. So upper and lower sums can get as close as desired, and this function, with one jump discontinuity, is integrable.
  3. Can a function with infinitely many discontinuities be integrable? Well, consider f(x)=1 if x=1/n for any positive integer n and f(x)=0 otherwise. Then f is integrable on [0,1]. Take a partition [0,ε,now sprinkle points around the finitely many discontinuities very closely]. Then the interval [0,ε] contains infinitely much of the trouble, and the total difference between the upper and lower sums is ε, really under control. The other finitely many discontinuities can be handled as in the previous example. So a Riemann integrable function can apparently be discontinuous at infinitely many (at least countably many) points.
    In fact, the function f(x)=sin(1/x) for x>0 and f(0)=anything is Riemann integrable on [0,1]! This is because we can put a box around 0 of height 1 and width ε, and everything is good outside, and the error, inside the box, with all the wiggling ("discontinuity of the second kind") is controllable.
  4. If f(x)=1 for x rational and f(x)=0 for x irrational (this is "the characteristic function of the rationals"), then f is not Riemann integrable on [0,1]. This is because in any interval of positive length, there are rationals and irrationals (density) and therefore the upper and lower sums will always differ by 1.
  5. A more subtle example is the following. Suppose f is the characteristic function of the Cantor set. So f(x)=1 when x is in the Cantor set, and f(x)=0 otherwise. This f surprisingly is Riemann integrable and its integral is 0. I remind you that x is in the Cantor set if the ternary expansion of x has only 0's and 2's. And the Cantor set is constructed by throwing our middle third intervals. So at the nth stage, we have [0,1]=Inn∪Outn. There set Outn is 1+2+22+...2n open intervals, and the total length of Outn→1. So we can enclose Inn inside intervals whose total length is as small as we like by taking n very large. Use the endpoints of those intervals to make a partition. The upper and lower sums of this partition will be as small as we like.
The function is discontinuous on the Cantor set, and the Cantor set has uncountably many points. So the cardinality (the size, as a set) of the collection of discontinuities seems not to matter. What matters is how the set of discontinuities is put together on the line, a geometric (or, rather, metric) property. There is a rather surprising and, in its way, definitive result proved by Henri Lebesgue near the beginning of the 20th century.

When is a bounded function Riemann integrable?
Suppose f:[a,b]→R is bounded. Then f is Riemann integrable if and only if the collection of discontinuities of f has measure 0.

The function f(x)=1/n if x is rational and m/n in "lowest terms" and 0 otherwise is Riemann integrable. Its set of discontinuities is the rationals. Given ε>0, you can, with some ingenuity, construction partitions with this function's lower sum=0 and its upper sum smaller than ε. The characteristic function of the rationals (1 on the rationals and 0 otherwise) is not Riemann integrable. Its set of discontinuities is all x and this is not a set of measure 0.

A set of real numbers has measure 0 if and only if, given any ε>0, the set is inside a countable union of intervals whose total length is less than ε. Professor Anton Schep of the University of South Carolina, has a very nice complete proof, with sufficient background to be understandable of Lebesgue's result. It is not very long (with introductory material, only 3 pages). I was wrong in class when I stated that this result is not in the textbook. A version appears in Chapter 11, far, far away.

Of course Lebesgue published his theorem as part of his complete overhaul of the theory of integration. My online dictionary gives for overhaul

   a. take to pieces in order to examine.
   b. examine the condition of (and repair if necessary).
Here is an extremely simple example of what does go wrong. Suppose f is the characteristic function of the rationals. It is 1 for rational numbers and 0 for irrationals. Then ∫01f doesn't exist. But define the function fn by fn(x)=1 if x is the nth rational (assume you have some enumeration in mind here, please) and fn(x)=0 otherwise. Then since fn has only one little discontinuity, it is Riemann integrable (with integral equal to 0). But f=∑n=1fn clearly. So the Riemann integral doesn't work well with infinite sums. It needed fixing. In fact, there are many ways to fix it. There's some discussion of this in David Bressoud's books, mentioned here.

Monday, November 17, 2008

All from the book. What! Not in the book?
So here is a sort of differential equation which has no solution:
      f´(x)= 1 if x>0 and 0 if x≤0.
This is because the specified derivative does not have the Intermediate Value Property. But such functions have occurred as parts of models of reality, as some engineers and physicists see reality. And, in fact, we have a course, Math 421, which discusses how to solve such equations, in spite of the fact that they apparently have no solution. Indeed, any child of 5 can tell you that       f(x)= x if x>0 and 0 if x≤0.
The technique that's taught in 421 is called the Laplace transform. An early champion of this method, using the phrase "operational calculus", was Oliver Heaviside who wrote, Should I refuse a good dinner simply because I do not understand the process of digestion?

It should be true that the function which is piecewise defined to be 0 on (-∞,0) and x on (0,∞) should have derivative 0 on the left and 1 on the right. What happens at 0 shouldn't be very important. Now people declare that there is no classical solution whose derivative is 0 on the left and 1 on the right. There are notions of generalized solutions and distributional derivatives. This was all systematized in the 30's and 40's by Sergie Sobolev and Laurent Schwartz, more or less working independently. One quote in Schwartz's biography is interesting: To discover something in mathematics is to overcome an inhibition and a tradition. You cannot move forward if you are not subversive.

Let me try to explain the idea of a distributional derivative in the simplest case. Suppose f and g are functions on R1 and f is differentiable with derivative g, and g is a continuous function. Then we write f´=g, of course. We know from our discussion last time that there are many smooth (even C) functions φ(x) whose support is compact. The support of a function is the closure of the set where the function is not 0. So, for example, the function which is 1 on (0,1) and 0 elsewherer has support equal to [0,1]. But we are interested in functions φ which are smooth and have compact support. Now look:

Step 1 Since f´(x)=g(x), of course φ(x)f´(x)=φ(x)g(x).
Step 2 We integrate: ∫-∞ φ(x)f´(x)dx=∫-∞φ(x)g(x)dx. Now it may seem that these are improper integrals, because of the appearance of all the ∞'s. But remember that φ will be 0 outside of [-A,A] if A is a large positive number since φ has compact support and compact sets are closed and bounded.
Step 3 Consider the left-hand side of the equation: ∫-∞φ(x)f´(x)dx. It is actually ∫-AAφ(x)f´(x)dx for some large A where φ(-A)=0 and φ(A)=0. Now (clever, clever, clever!) integrate by parts:
     ∫-AAu dv=uv]-AA-∫-AAv du
This throws the derivative on the "other factor" with the penalty being the boundary term (with the ]) and a minus sign. Here we will take u=φ and v=f´. Notice that the boundary term, when both A and -A are "plugged in", must be 0 because of the assumption about the support of φ. Therefore ∫-∞φ(x)f´(x)dx is exactly the same as –∫-∞φ´(x)f(x)dx.
Step 4 We have shown that if f´=g as classical functions, with the classical derivative, then

So how to define the derivative if a function is not differentiable?
Now as a result of the sequence of steps above, we have a consequence of f´=g which doesn't have a derivative of f! More importantly, it turns out that the consequence is reversible. That is, if f has a continuous derivative, g is a continuous function, and if the equation is true for all smooth φ's with compact support, then f´=g. So now let's just support that f and g are continuous. We will say if the equation is true for all smooth φ's with compact support, then g is the weak or distributional derivative of f. This turns out to serve wonderfully. There is a tremendous amount of analytical (and algebraic!) intricacy in this defintion, so it needs to be considered in detail later.

Back to the text ...

The original statement of L'H, which was first proved by Johann Bernoulli, appeared in a very early calculus book authored by Guillaume François Antoine Marquis de L'Hôpital (the Wikipedia biography is shorter and more fun). The proof there mostly consists of a diagram showing two curves crossing on the x-axis together with their tangent lines there, and an accompanying discussion, more or less declaring, "There, see it!" I am sure that learning a rigorous proof is useful. I can't internalize the rigorous proof. Oh well. This is one of the standard proofs and is the proof, I believe, used in our current calculus text.

Thursday, November 13, 2008

Definition of derivative

Differentiability implies continuity. Algebraic combinations of differentiable functions are differentiable. xn is differentiable (when n is a positive integer).

The Chain Rule
Proved very slickly by stuffing one linear approximation inside another and realizing that the consequence was, indeed, the Chain Rule.

The beginnings of a gallery
Everyone, young and old, learned and naive, should have a gallery of functions to be considered when theorems and definitions are considered. The instructor spent a chunk of time identifying good initial candidates for such a gallery, from "ugly" to "nice". Of course, the beauty of the qualities mentioned depend quite a bit on the observer! Writing in progress!

  1. The inverse (?) ruler function
  2. The characteristic function of the rationals
  3. The ruler function (Dirichlet?)
  4. x sin(1/x)
  5. x2sin(1/x)
  6. Bumps, etc. (due to Emile Borel)
  7. The exponential function

Monday, November 10, 2008

Increasing functions
I discussed increasing functions and how often they can be discontinuous. This is in the textbook. I wrote and gave a warm advertisement for a book: Functional Analysis by F. Riesz and Nagy (also known and pronounced as "rees-naje". Reprinted by Dover and on sale at Amazon for 14.93. A wonderful book!

Mr. Skalit
Discussed several problems in chapter 4 he had prepared with the help of Mr. Kowalick. In particular, we saw that metric spaces were normal.

The instructor defined and began a discussion of differentiability and announced his preference for a definition written in this form: f:[a,b]→R is differentiable at c∈[a,b] with derivative Q if there is a function E defined in a neighborhood of 0 with limv→0E(v)=0 so that f(x)=f(c)+(Q)(x-c)+E(x-c). This definition doesn't have any divisions.

Some simple consequences were mentioned but the Chain Rule will be verified next time.

An exam was announced, to be given in two weeks. Ms. Hood volunteered (??!) to present a problem about differentiation, with the help of Mr. Baldwin.

Thursday, November 6, 2008

The due date for the next homework assignment is Monday's class meeting.

Finishing the proof
We used compactness to verify uniform continuity. I continued with the attempted (or rather, interrupted) proof. So X is compact, and we have sequences {pn} and {qn} with d(pn,qn)<1/n and d(f(pn),f(qn))≥ε. Now {pn} is a sequence in a compact metric space. The sequence itself may not converge, but compactness always guarantees that the existence of a convergent subsequence, {pnj}. We know that there is an x so that limj→∞pnj=x. Notice that the sequence {qn} is sort of "dragged along" by the corresponding sequence of pn's. That is, I claim the subsequence {qnj} also converges, and its limit is x. Why is that? Well, d(qnj,x)Δ≤d(qnj,pnj)+d(pnj,x). The first term on the right is <1/nj. For j large, this will be small and stay small. And the second term behaves similarly, because the p-subsequence converges to x. Now f is continuous at x. Therefore given ε>0, we can find a δ>0 so that if d(x,y)<δ then d(f(x),f(y))<ε. But choose J so that for j≥J, both d(qnj,pnj) and d(pnj,x) are less than δ/2. Then d(qnj,x)<δ so d(f(qnj),f(x))<ε and d(f(pnj),f(x))<ε. Again the triangle inequality applies, so d(f(pnj),f(qnj))<2ε for j>J. This is a contradiction to how we created the two sequences initially (that is, it will be a contradiction if you allow me to relabel 2ε as ε!). So we are done.

A consequence of uniform continuity?
I proved a rather remarkable result, and this will be especially wonderful in many applications later. I mentioned that a "random" (?) continuous real-valued function on R can be rather bizarre. One explicit well-known example is xsin(1/x) on [0,1]. For another example with even worse behavior, I investigated the Takagi function graphically in the 9th meeting of the Byrne Seminar on Experimental Math. A picture drawn there is shown to the right. The Takagi function (named for Teiji Takagi (1875--1960)) happens to be an example of a continuous nowhere differentiable function on [0,1], and it is also a function which is neither increasing nor decreasing on any subinterval of [0,1] of positive length. So continuous functions can be weird and wonderful. In view of that, the result here is almost amazing. If you are willing to tolerate only a little bit of error (as little as you would like) you can approximate any continuous function by a piecewise constant step function with finitely many values. David Tall, an English mathematics educator, gives a detailed discussion of the properties of this function (aimed at teachers) here.

To the right is a possible picture. The graph of a continuous function defined on a closed bounded interval is shown. The light blue band around the graph is ±ε deviation from the graph. The red horizontal lines are the graph of an approximating piecewise constant function with finitely many values. It "steps" because the values it assumes are taken on finitely many intervals of positive length. The result is remarkable to me, because no matter how bad (?) the function f can be, we can replace it by something with finite data (!) if you allow me to commit a very small error. In many applications, the step function is too coarse, and we may want piecewise linear or differentiable or ... whatever. But this result is the initial version. So:

Theorem Suppose f is a continuous real-valued function on [a,b], and ε>0. Then there is a function g:[a,b]→R so that the range of g is finite, g-1(y) is either empty or a finite set of intervals for all y∈R, and |g(x)-f(x)|<ε for all x∈[a,b].
Proof [a,b] is compact and f is continuous, so f is uniformly continuous on [a,b]. Therefore we can find δ>0 so if |f(x1-x2|<δ, then |f(x1)-f(x2)|<&epsilon. Now find a positive integer n so |b-a|/n<δ. For all integers j between 0 and n-1, we know that |f(a+j(b-a)/n)-f(x)|<ε for x between a+j(b-a)/n and a+(j+1)(b-a)/n. Define g by this "rule": if x is between a+j(b-a)/n and a+(j+1)(b-a)/n, g(x)=f(a+j(b-a)/n). Then we're done!
Advertisement This is magic, and you should realize it. We have developed a considerable amount of technique, and now many results, including what's in this lecture, are "easy" to verify. Once we have the step function, we can (approximately) integrate f, find its (approximate) mean and standard deviation, etc.

d) implies our first inverse function theorem
There are a collection of results which are referred to as inverse function theorems, including, especially, a result about differentiable mappings between open subsets of Rk. This wonderful result will be proved in 412. One of its standard proofs uses the following result, which is also very frequently invoked in algebraic topology. So: Theorem Suppose f:X→Y is a bijection (1-1 and onto -- basically, X and Y are the same as sets!, f is continuous, and X is compact. Then f–1 is continuous.
Proof We will use d). To show that f–1 is continuous, we will investigate if (f–1)–1=f has the following property: if C is closed in X, then f(C) is closed in Y. But X is compact, so C, a closed subset, is compact. f(C), a continuous image of a compact subset, is compact in Y. But compact subsets of metric spaces are closed, so f(C) is closed in Y. And we are done!
Advertisement This is magic. There is hardly any effort to verify a result which is used constantly. Be aware, if you know what a topological space is, that the result is true if only X is a compact Hausdorff space.

As I mentioned, NJ State Law QD 324-17 (a math regulation) requires that this example be presented immediately after the statement and proof of the previous theorem.

Consider [0,1) with its usual topology. This is 1 point away (?) from being compact. The mapping f:[0,1)→R2 given by f(t)=(cos(2πt),sin(2πt)) is continuous. Considering the picture is more fun. The range of f is x2+y2=1, a closed and bounded subset of R2, so the range is compact. But f–1 is not continuous. Why? The picture shown to the right gives some idea. Take the magenta colored dots in the range, which approach the image of 0 from the wrong way. These dots have a limit, the image of 0. But pulled back to [0,1), we get the green dots which have no limit in the domain of f. Thus f–1 takes a convergent sequence to a sequence which does not converge, and a continuous function can't do that.

A major result used in calculus
If f:[a,b]→R is continuous, then f([a,b])=[c,d]. This result includes the Intermediate and Extreme Value Theorems. The proof is (now!) easy. A proof from just the definition of continuity is likely to be tedious. So why is this result true? [a,b] is compact and connected, so f([a,b]) is compact and connected. Connected subsets of R are intervals. The only compact intervals are those which are closed and bounded. So we're done.

As mentioned, the lecturer thinks that connectedness as defined is a difficult concept. The definition has a not very prominently mentioned, so it may be irritating to verify that something is connected (or not connected). There is a negative (?) aspect about the logic. So here is a variant.

Arcwise connected
A metric space X is arcwise connected or pathwise connected (both phrases are used) if for all pairs of points p and q in X, there is a continuous function c:[0,1]→X so that c(0)=p and c(1)=q.
Theorem If X is a pathwise connected metric space, then X is connected.
Proof If X is not connected, then X=A∪B with A and B open, non-empty, and disjoint. Take p∈A and q∈B. Then pathwise connected provides a continuous c as in the definition. Since 0∈c–1(A) and 1∈c–1(B) and c is continuous, then (using c)!) these are open subsets of [0,1], disjoint, non-empty. But [0,1] is connected, which is a contradiction.

Star-shaped subsets of Rn are arcwise connected, because a star-shaped set has a center, v (or, at least one center!), so that given any p in the set, the line segment from p to v is in the set. So we can connect p to q by detouring (?) through v. Convex sets are star-shaped. Open balls are therefore star-shaped.

For open subsets of Rn...
Much, although not all, of analysis in 411-412 will be in open subsets of Rn. For such subsets, the concepts of arcwise connected and connected coincide.
Theorem Suppose U is an open subset of Rn. Then U is connected if and only if U is arcwise connected.

Proof I did not offer a proof in class, but look: suppose U is connected. I want to show that U is arcwise connected. Fix p∈U. Take A to be the set of all points q which can be "connected" to p with a continuous image of [0,1], as in the definition of arcwise connected. Then A is a connected set, surely. I claim that A is open. Well, if q∈A, then since U is open, there is r>0 with Nr(q)⊂U. I can connect from p to any point in Nr(q) with a detour through q. So A is open. Also, A is closed, since a limit point of q's, call it v, has a ball Ns(v)⊂U. At least one q is in Ns(v), so go from p to that q and then, inside the convex ball, to v. Since A is open and closed and non-empty, A must be all of the connected set U. (Or else what is left out is open, also, and makes U disconnected!)

Not the other way!
For unfriendly sets, arcwise connected and connected may be different. For example, the topologist's sine curve is connected but not arcwise connected. There is a purported proof in the linked page, without much details and also without a picture! Here is a Wikipedia reference, with a picture, but not even a candidate for a proof.

I sketched a rather clumsy proof. My effort would involved repeated use of the Intermediate Value Theorem.

Monday, November 3, 2008

Definitions (?) and theorem about continuity
Here are a bunch of equivalent statements about a function f:X→Y, where X and Y are metric spaces.
a) For all p∈X, with f(p)=q, and for all ε>0, there is δ>0 so that if d(x,p)<δ then d(f(x),q)<ε.
b) If {xj} is a convergent sequence in X with limj→∞xj=p, then {f(xj)} is a convergent sequence in Y with limj→∞f(xj)=f(p).
c) If U is open in Y, then f–1(U) is open in X.
d) If C is closed in Y, then f–1(C) is closed in X.

All except the last were previously shown as equivalent. The last is an easy consequence of set and function manipulation results, using fact that a set is closed exactly when its complement is open. The great freedom we now have is to investigate and diagnose results about continuous functions (those which satisfy any/all of the previous properties) and use whichever characterization we like. Some are more convenient to use than others in certain situations.

Composition of continuous functions
We're given f:X→Y and g:Y→Z both continuous. Use c). Take U open in Z, then g–1(U) is open in Y, and f–1(g–1(U)) is open. Now notice that (gof)–1(U) is the same as f–1(g–1(U)).

Distance as a continuous function
We saw that d(x,y1)–d(x,y2)≤d(y1,y2) because (Δ≤) d(x,y1)<d(y1,y2)+d(x,y2). Then, switching y1 and y2, we know |d(x,y1)–d(x,y2)|≤d(y1,y2). Therefore if f(y)=d(x,y), f satisfies a Lipschitz estimate, and there must be continuous. See what follows, please.

Lipschitz and locally Lipschitz
A function f:X→Y is Lipschitz if there is a constant, A≥0 so that d(f(x1,x2)≤A d(x1,x2) for all x1, x2 in X. Such a function must be continuous. Use, for example, a). Given ε>0, take δ=ε/A (hey, talk to me if A=0 and you can't figure out what to do).
We'll say f:X→Y is locally Lipschitz if it is Lipschitz in a neighborhood of every point. That is, given x∈X, there is r>0 and A≥0 (A may depend on r and x) so that for all x1, x2 in Nr(x), d(f(x1,x2)≤A d(x1,x2). Such functions are also continuous (same proof!).

Lipschitz conditions are fundamentally important in the standard statement of the existence and uniquenss theorem of ordinary differential equations.

MVT implies Lipschitz
This is jumping ahead a bit, but if f:R→R is differentiable, then the Mean Value Theorem implies that |f(b)–f(a)|≤|f´(c)| |b–a| for at least one c in the interval between a and b. Therefore if we "happen" to know that f´ is bounded on that interval, then f will be Lipschitz there, and we know a Lipschitz constant (the sup of the absolute values of the derivative). Of course, a differentiable function will automatically be continuous, but this maybe gives some feeling for what Lipschitz means: the mapping "stretches" distances by no more than a factor of A, the Lipschitz constant.

Hölder and locally Hölder
Here is a similar idea. Suppose A≥0 and α≥0. Then f:X→Y satisfies a (uniform) Hölder condition if d(f(x1,x2)≤A (d(x1,x2))α. For example, the function f:R→R defined by the formula f(x)=sqrt(|x|) satisfies such a condition with α=1/2. It does not satisfy a Lipschitz condition on all of R.
An analogous definition can be made for locally Hölder.

Functions satisfying Hölder conditions are also continuous. For example, take δ=(ε/A)1/α. The "average" continuous function on [0,1] is not differentiable, but I think it does satisfy, in most places, a Hölder condition of order 1/2. So this is important in many fields (e.g., probability, math finance, physics [Brownian motion], etc.).

Algebraic combinations
Sums, products, and quotients of complex-valued continuous functions are continuous. Probably the simplest way to verify this is by appealing to similar statements about convergent sequences and using the criterion in b) above.

Vector-valued functions and their components
Suppose F=(f1,f2,...,fk):X→Rk. Then each component, fj, is a function from X to R. And F is continuous if and only if each of the fj's is continuous. The easiest way to see this is probably to contrast the Euclidean metric, L2, in Rk with the L1 and L metrics, and use the constants we have previously obtained in problem #3 of Homework #2 to show the needed implications. This is also in the text, of course.

Perturbing the definition of continuous
Suppose f:X→Y. A version of the official definition of continuity follows:

∀p∈X∀ε>0∃δ>0 if x∈X with d(x,p)<δ then d(f(x),f(p))<ε.
A wonderful thing you could do to yourself is change the order or the type (∀ to ∃ or ∃ to ∀) of the quantifiers and see what that does to the definition. For example:
∀p∈X∃δ>0∀ε>0 if x∈X with d(x,p)<δ then d(f(x),f(p))<ε.
I think such functions still are continuous, but additionally there aren't "very many" of them -- they are functions which are locally constant. That is, any function at any point has a neighborhood where it is constant. You might want to prove that such a function on a connected metric space must be actually constant.

What about this?
Here's another change to the definition.

∀ε>0∃δ>0 if x,p∈X with d(x,p)<δ then d(f(x),f(p))<ε.
These functions are also continuous. But does every continuous function satisfy this statement? The statement is more mysterious, since, given ε>0, we're making a selection of δ>0 which will "work" for any pair of points in X.

An example
An instructive example is f:R→R defined by f(x)=x2. We can show that this function does not satisfy the previous logical statement. We decided that we needed to produce an ε>0 so that for any δ>0 there are points p, q in X with d(p,q)<δ with d(f(p),f(q))<ε. Then we quickly declared that this would be proved if we choose ε=1, say, and created, for each positive integer n, points pn and qn so that |pn–qn|<1/n but |(pn)2–(qn)2|≥1.

The marvelous intervention of Ms. Slusky allowed us to choose pn=n and qn=n+(1/[2n]). We computed |(pn)2–(qn)2| and saw that it was greater than 1. The instructor then drew a picture intending to show that the graph of this f tilted more and more as |x|→∞ so that getting control over |f(x)–f(y)| even if the size of |x–y| was restricted became more difficult. For more pictures, please see the link below.

A definition
f:X→Y is uniformly continuous if

∀ε>0∃δ>0 if x,p∈X with d(x,p)<δ then d(f(x),f(p))<ε.
Such functions are extremely important in applications, and even in such basic situations as integration. Please see a digression (or discussion) in Math 503 about uniform continuity in the basic complex variables graduate course.

Some other examples, perhaps elaborate
Example 1 x2 is not uniformly continuous on R.
Comment about learning stuff Especially in math, when the defintions get more and more complicated, a very good idea is to find things which do not satisfy the definition yet are very closely related to things which do satisfy the definition. The differences may be useful in understanding the definition and its consequences.

Example 2 MVT again: conside the function f:R→R defined by f(x)=(3/[4+x2])+17x+cos(4x). I claim that f is uniformly continuous. I'll show this by checking that f satisfies a uniform Lipschitz condition. Then the same δ can be used for any ε and any x. Well, f´(x)=–(6x)/[4+x2]2+17–6sin(x). I claim that each of the "pieces" of this derivative is bounded in all of R. That should be clear fro 17 and –6sin(x), so let's look at (absolute value!) 6x/[4+x2]2 for x>0. Uhh ... this is less than 1. Why?
6x≤[4+x2]2: true for x between 0 and 1, certainly, since the largest that 6x gets there is 6, and the smallest that the right-hand side gets is 16. And consider the derivatives:
6≤2([4+x2]2x. The right-hand side is always at least 2(5)2=20 on [1,∞) so it is always larger than 6. Therefore the original functions have the same inequality on [1,∞). This is too much work! It should be easier.

Example 3 5sqrt(x)+17x on [0,∞). Well, the derivative is 5/(2sqrt(x))+17. Mr. Leven correctly objected here because the derivative is not bounded on [0,∞). O.k.: the derivative is bounded on [1,∞) so the function is uniformly continuous there. What about on [0,1]. Aha! Look ahead.

A major theorem
If f:X→Y is continuous and X is a compact metric space, then f is uniformly continuous.

Attempted proof
There's a nice proof in the textbook, using essentially problem 5a) from the first exam. Let me try to give a different proof. What if the theorem were false? Then we would have some ε>0 and two sequences of points {pn} and {qn} (remember the verification of the x2 example above) so that d(pn,qn)<1/n but d(f(pn),f(qn))>ε. This alone is not enough to guarantee a contradiction, because we could have such sequences in R and not get into trouble. (In class I gave sequences in R2, pn=(0,n) and qn=(0,n+{1/n}) but we already had the examples in the x2 verification.) What will help us here?

Hint Compactness! This guarantees not that a sequence itself converges, but that some subsequence does. This will be enough to get a contradiction. to be continued ...

Thursday, October 30, 2008

Products of absolutely convergent series
The ingredients (hypotheses) include the following:
Then (the conclusion!) the sequence {st} converges to AB.

I proved this in detail, perhaps too much detail. The process used the convergence definition applied to the infinite series ∑j=1aj and the infinite series ∑k=1bk, and the Cauchy criterion applied to ∑j=1|aj| and to ∑k=1|bk|. The proof used the "technique" of problem 7 in the first exam, and was guided by a version of the rather bizarre picture to the right.

Love for absolute convergence
The rearrangement result and the product summation result declare that commutativity and associativity are correct when applied in (essentially) any way to absolutely convergent series and to their algebraic manipulations.

Other results about products
I stated Merten's Theorem and another result. Please see the text.

Limits and sequences
I stated the ε–δ definition of limit, and proved that it was equivalent to the sequential statement. This is as in the text. Mr. Baldwin kept me honest.

I defined continuity of a function at a point, if the function mapped one metric space to another. Then I defined continuity of a function. I showed that this definition ("Inverse images of open sets are open") was equivalent to requiring an ε–δ statement at each domain/range point pair. This is as in the text. All is an abstract version of the result on real-valued functions defined on the real line which was stated several weeks ago.

Ms. Pritsker suggested that I try to show some pictures about what was happening, and I thank her for this.

In this first example, the function shown (from R to R) has what's called a jump discontinuity. The inverse image of the green open interval is traced (backwards) and the result seems to be a half-open interval. So this example fails the "inverse image open" criterion for continuity.
The second example has a more complicated discontinuity. I tried to sketch, in class and here, the function f whose value, f(x), 0 if x≤0 and is sin(1/x) if x>0. Then the inverse image of a small open interval centered around 0 includes (–∞,0] and also includes a countable collection of open intervals, getting smaller and smaller as they "pile up" at 0. Note, though, that although 0 is in the inverse image, there is no open neighborhood (consider the right "half") which is a subset of the inverse image.

The great theorems
Chapter 4 concentrates on precise statements and proofs of some of the most basic theorems in analysis about continuity. These results seem to have their historic source (sort of) in the work of Bolzano. The calculus version of the theorems is the following almost ludicrously simple statement:

Suppose f:R→R is a continuous function, and a≤b. Then the set of values of f([a,b]) is [c,d] with c<d.
This result of course includes the Intermediate Value Theorem and the Extreme Value Theorem. By the way, the converse of this result is not, I believe, correct (that is, continuous functions are not the only functions from R to R which obey the conclusion of this theorem).

Notice that closed bounded intervals in R can be characterized as proper non-empty subsets of R which are both connected and compact. We will prove the result above by showing that continuous images of compact sets are compact, and continuous images of connected sets are connected. Sigh. Both compact and connected are defined in terms of open coverings. So it turns out that the "inverse image open" version of the definition of continuity is exactly suited for quite simple proofs of the results needed. But, please, the apparent simplicity is only superficial. The whole theory is aimed at these results, and 150 years of work have gone into constructing a Definition/Theorem/Proof/Example succession which seems simple and nearly effortless. Historically the results and the ordering involved a great deal of work.

Please read chapter 4. The top vote-getters in the poll for "Problems students want to do" are Chapter 3: 6, 10 and Chapter 4: 4, 6, 7, 11. I will add one more problem in a formal assignment on Monday.

Monday, October 27, 2008

Rearrangements of absolutely convergent series
We proved this result. How did we do it? Well, we know that ∑j=1|aj| converges. And we let A=∑j=1aj, since the original series must converge. And σ:N→N is the rearrangement. So we need to analyze the partial sums of ∑j=1aσ(j) and look at their differences with A. We used both absolute convergence and the Cauchy criterion for the series with absolute values.

An example
Here is one (delightful?) example of the sort of computation which made people very uneasy historically. You may remember from the first year of calculus study (geometric series, Taylor series, remainders, etc.) a fact about the Alternating Harmonic Series.

That the series converges follows from a version of the Alternating Series Test which we will verify in a few minutes. The specific value of the sum relies on things like Taylor's Theorem.

Now replace each positive term by twice itself minus itself. I hope you can convince yourself this does not change convergence or the sum of the series. Now this is true:

and now divide this series (and its sum) by 2. I mean (emphasis!) divide everything by 2. Here is the result:
It is not difficult to check carefully that what is apparently correct actually is correct. The series on the right-hand side is a rearrangement of the original Alternating Harmonic Series. Wow! This rearrangement converges to half of the original value. We don't really need to know what that value is, but it is (relatively) easy to see that the original value was actually positive. So rearrangements of this series do change the sum!

Riemann Rearrangement Theorem
If a series of real numbers converges yet diverges absolutely (conditional convergence) then there is a rearrangement which converges to any real number. I gave a proof of this. This is a weaker version of the result in the textbook. Please look there.

The better version in the text
Rearrangements can be devised to have the sequence of partial sums with lim sup and lim inf very arbitrarily specified. There are versions of the theorem which apply to complex series. The conclusions then are more complicated, though.

Summation by parts
We went through the discussion of summation by parts. Several students clearly understood it better than the instructor. Sigh. This was discussed because summation by parts can be applied to several topics in Fourier series.

Some consequences (Alternating series test)
We stated and proved the standard alternating series test.

Products of series
Here we began a discussion of products of series, which is a more complicated undertaking than is immediately apparent. We want to start with two convergent infinite series, ∑j=1 (or 0)aj and ∑k=1 (or 0)bk and then analyze the numbers ajbk. We'd like to figure out some way to "assemble" these numbers into an infinite series, whose sum, we might hope, is AB, the product of the sums of the two series we began with.

Taylor series and Cauchy products
Thinking about Taylor series almost immediately inspires (?) the definition of the Cauchy product. We "know" from calculus ("know" means that some examples were shown then!) that if f(x)=∑j=0ajxj and g(x)=∑k=0bkxk then a way (?) to organize the product is to suppose it is F(x)=f(x)g(x), a product of functions. And we would hope that F(x)=∑t=0ctxt where now ct=∑q=0taqbt–q. These formulas are gotten in several ways: first, we rewrite the product by assuming that manipulations with infinite polynomials work exactly like those with finite polynomials. Or we can hope that a Taylor's Theorem is at work behind the scenes, and the summation exactly reflects the product rule for lth derivatives. The sum is sometimes called a convolution of the two sequences. It is unfortunate or, perhaps, interesting, that many of these hopes are not exactly true all of the time we would hope. Oh well. In particular,here is no clear relationship between convergence of the factors and convergence of the Cauchy product. See the example mentioned below.

Example in the text
A neat example in the text shows that the Cauchy product of a convergent series (with itself!) need not converge. Interesting. I have little to add to the account in the textbook.

Dirichlet series and Dirichlet product
Here I defined Dirichlet series, which are rather useful in number theory, and some other areas which don't seem to be immediately related. These are series of the form f(s)=∑n=1(an/ns). Such series are natural in number theory and complex analysis. The most famous example occurs when all of the an's are 1. This is ζ(s)=∑n=1(1/ns), the Riemann zeta function. Learn more about this, and make a milliion dollars!. Dirichlet series have a whole theory of their own. For example, just as power series have a radius of convergence inside which they converge absolutely, Dirichlet series similarly (and for much the same reasons!) have an abscissa of convergence, a line x=Constant in the complex plane. The series converges absolutely to the right of that line and diverges to the left of it. For ζ(s), that line is x=1. Or, as we say in the complex analysis game, Re(s)=1.

If we have another Dirichlet series, say g(s)=∑m=1(bm/ms), we might consider F(s)=f(s)g(s) and think about ∑n=1(an/ns)·m=1(bm/ms), and maybe reassemble into another Dirichlet series, so: ∑N=1(cN/Ns) and this should be the same as ∑n=1m=1(an/ns)(bm/ms). So we might want cN/Ns to be, well, which terms of the product? To match up (1/ns)(1/ms) with 1/Ns we need nm=N.

Therefore we could define the Dirichlet product of two series (dropping the s stuff) ∑n=1an and ∑m=1bm to be the series ∑N=1cN where cN=∑n·m=Nanbm. Here it is divisors which are important. You can see if factoring integers and primality is interesting, such products might be more revealing than Cauchy products. One can then ask if convergence of the factors always implies convergence of the Dirichlet product. The answer, with no further hypotheses, is no. There are examples which don't have the desired behavior.

Absolute convergence and products
We will briefly discussion summation methods for products of series. Let me start with 1 in these series, and starting with 0 would give a similar result. So let's have two series ∑j=1aj and ∑k=1bk which converge and which have sums A and B respectively. Consider the numbers ajbk which I could think of as "sitting" each at a lattice point, NxN, of the plane. Now we will consider a sequence of finite subsets {Wt}t=1 of NxN, with the following properties:
     1. They are nested: Wt⊂Wt+1.
     2. t=1Wt=NxN: the union is "everything".

Examples of such Wt's come from both Cauchy and Dirichlet products. Adding up the terms which come from the Wt's exactly will correspond to the partial sums of each of these products. The easy and natural result (I love absolute convergence!) is the following.

Theorem Suppose that the series ∑j=1aj and ∑k=1bk both converge absolutely and have sums A and B respectively. Then the sequence of (possibly complex!) numbers st=∑(k,l)∈Wtajbk converges, and its limit is AB.
Actually, even slightly more is true and follows from the preceding result. That is, the sequence whose tth term is ∑(k,l)∈Wt|ajbk| also converges, and its value is ≤ the product of ∑j=1|aj| and ∑k=1|bk|. So any method of summation of the product series of absolutely convergent series "works" and gives the correct answer.

We began thinking about the proof and will complete it next time.

A volunteer and his coach
Mr. Skalit, helped by the valiant Mr. Kowalick, will present problems 20, 21, and 22 of chapter 4 in class, probably a week from Thursday.

Problems to be considered
The students in the class will consider Chapter 3: 6, 7, 10; and Chapter 4: 4, 5, 6, 7, 11, 13, 25. Votes for the best problems will be counted, and the 6 or maybe 7 highest vote-getters will be assigned as homework, due on Thursday, November 26.

Thursday, October 22, 2008

So we quickly reviewed some series material after the traumatic first exam. The instructor proved the Root Test as in the text, and them proved the famous Cauchy-Hadamard Formula for the radius of convergence of a power series. This led naturally, through the investigation of the power series of what turns out to be the exponential function (Stirling approximation of factorial) to the Ratio Test, some parts of which were proved. Then we discussed the folklore that, If the Ratio Test applies, then the Root Test applies. (and they'd better give the same answer!) We proved this, even going over part of the proof which is not in the text (!).

The article I mentioned in class which taught me new things about the Ratio Test is this: The mth Ratio Test: New Convergence Tests for Series and is written by Sayel A. Ali. It appears in the June-July 2008 issue of the American Mathematical Monthly.

I started to discuss Why I love absolutely convergent series! My first reason was that rearrangements of such series also converge, and converge to the same sum. This is nice. I didn't quite finish the proof, and I will give an interesting example next time, and a result of Riemann's which explains what happens when the series does not converge absolutely.

Thursday, October 16, 2008

There was little added value by the lecturer, who discussed the convergence of some special sequences. However, Ms. Sergel, assisted in her preparation by Ms. Slusky, gave a really nice discussion of the convergence of the very classical algorithm for sqrt(x) (a recursive sequential algorithm known in China and India for about 2500 years, and introduced in Europe only about 500 years ago). This is an algorithm which is very widely used, and the analysis of its convergence is elementary and inspired.

The lecturer continued in a more pedestrian fashion. He defined infinite series. He talked about translating various sequence statements to infinite series: the Cauchy criterion, series with non-negative terms, absolute convergence, comparison, p-series via the Cauchy condensation theorem, and a brief anticipation of a version of the ratio test. More next week!

Monday, October 13, 2008

Cauchy and complete again
We rapidly reviewed the results of the Cauchy and completeness discussion last time. Amazingly, students were either apathetic or knew better than to ask for further details about any of the proofs. Sigh.

Inequalities and limits
The only Euclidean space Rk where an order topology coicides with the usual Euclidean metric topology is R1, that is, R, the real numbers. Some special results follow from using the order in R. For example, we have this theorem:

Theorem Suppose {xn} and {yn} are two real convergent sequences with limits Lx and Ly, respectively. If for all positive integers n we know that xn≤yn, then Lx≤Ly.
A proof of this was shown, using contradiction. If Lx>Ly, then take Lx–Ly=2ε is positive. Take n large enough so that |xn–Lx|<ε and |yn–Ly|<ε. Then –ε<xn–Lx<ε and –ε<yn–Ly<ε. So we know that –ε<Lx–xn<ε (multiply the first inequality by –1) so that xn–ε<Lx<xn+ε. Similarly, –yn–ε<–Ly<–yn+ε so that (xn–ε)+(–ε–yn)=(xn–yn)–2ε<Lx–Ly<(xn–yn)+2ε=(xn+ε)+(–yn+ε). This is already a problem since Lx–Ly=2ε.

The instructor asked if a converse to this result was true, and was greeted with scorn. Indeed, no, since 0<0 but the first 0 could be the limit of the sequence {–1/n} and the second 0 could be the limit of the sequence {1/n}. Huh. But how about this:

Theorem Suppose Lx is the limit of the sequence {xn} and Ly is the limit of the sequence {yn}. If we know that Lx<Ly, then there is a positive integer N so that if n>N, xn<yn.

The proof of this is that (taking 2ε=Ly–Lx) if |xn–Lx|<ε and |yn–Ly|<ε again, then unroll the absolute values as before, you will see that the result quoted is true.

Notice that 1<2 and 1<5 but 1–1<2–5 means 0<–3 which is certainly false! Generally, you cannot subtract inequalities and be sure that the result is correct.

More subtle results occur, but we need a few more definitions.

∞ and –∞
Suppose {xn} is a real sequence which has the following property: for all M∈R, there is a positive integer NM so that if n≥ NM, then xn>M. Then we will say that limn→∞xn=∞.
If we changed "xn>M" in what's above to "xn<M" then we will write limn→∞xn=–∞.

lim sup and lim inf
Suppose {xn} is a real sequence. Define S to be the set of all subsequential limits of this sequence. This is a collection of numbers in R*, which I guess is R∪{∞}∪{–∞}, and we will think of R* as an ordered set with the (more or less) obvious ordering. I asked if we could find examples of sequences with S=∅ and S=R*. Well ...

How to think about this: a sort of bisection method. Now consider [–∞,∞] and split it into two parts, [–∞,0]\cup;[0,∞]. Look at N: let's call an integer n left if xn∈[–∞,0] and call it right. Hey, if it is both left and right call it both. In any case, N is now the union of a set of left and right sets, so one of them has infinitely many elements. Let's suppose it is the right one. Then consider [0,∞]=[0,1]\cup;[1,∞]. Hey: left/right again. If left triumphs (has infinitely many!) then split up [0,1] into [0,1/2]\cup[1/2,1] ETC.. If right wins, then split [1,∞] into [1,2]\cup[2,∞] ETC. Eventually we will get a finite limit or a subsequence which is pushed to infinity. In either case, S≠∞. Whew! So there must be at least one subsequential limit.

There is a sequence with S=R*. Take {xn} to be any "enumeration" of the rationals: so this refers to a function f:N→R which is 1–1 and whose range is Q. Then S is R*. Why? Because in any real interval of postive length there are infinitely many rationals, so given any such interval and given any integer J, there is always j>J with xj inside the designated interval. It is not difficult to use this observation to verify that S is indeed R*.

Question Is there a sequence which has S=R? You think about this, please.

lim sup and lim inf
Given {xn} consider S, the set of all subsequential limits. Then lim sup is the sup of S (considered as a subset of R*) and lim inf is the inf of S.

Some special sequences
The sentence "And now we begin." ends the novel, "Portnoy's Complaint", by Philip Roth. This sentiment could be echoed by many analysts at this point in the course. We now start looking at some concrete examples of sequences and then series.

The examples
We saw as a consequence of our study of the Archimedean property, that

The lecturer falls, and no student picks him up!
We tried to consider the sequence {a1/n} for a>0. This sequence does converge, and its limit is 1. After some confusion regarding the Bernoulli inequality, little progress was made towards the verification of the convergence and limit claim. Sigh. Deferred to Thursday's class, I suppose!

Thursday, October 9, 2008

We resumed with subsequences. The lecturer stated a theorem which Ms. Slusky observed was trivial. The lecturer was embarassed.

Writing in progress!
He continued with a clumsy proof that the set of subsequential limit points of a sequence in a metric space is always a closed set. There is a tiny bit of diagonalization needed in this proof. The lecturer obfuscated this as much as he illuminated it.

Cauchy sequences
A major discussion was started with the definition of Cauchy sequence, a very important concept in analysis and, indeed, in all of mathematics. Given {xn} in the metric space (X,d), we call this sequence Cauchy if given any ε>0, there is a positive integer Nε so that if n and m are integers ≥Nε, then |xn–xm|<ε.

Cauchy seuqences are candidates for convergent sequences. They are defined, though, only with "internal", self-referencing criteria, so that knowing such a sequence must converge (as it must, under certain circumstances!) is neat. In R and Rn Cauchy sequences must converge. The convergence of many algorithms is guaranteed using the Cauchy criterion (that is, by proving that the sequences produced using the algorithm are Cauchy). We looked at some examples in R and some conditions in R.
Example If {xn} is a real sequence and we know that |xn–xn+1|<1/n, then such a sequence need not converge and such a sequence need not be Cauchy. Here the example (reaching for our knowledge of calc 1) is xn=∑j=1n1/j, the sequence of partial sums of the harmonic series. This sequence does not converge (hey: x2n≥1+[(n–1)/2] with an easy argument, so the sequence is not even bounded). Why is this sequence not a Cauchy sequence? Well, we need to compare xn and xm when n and m are both "free" and large. If we make n large with m>n, the triangle inequality will only provide a bound like this:
|xn–xm|≤|xn–xn+1|+...+|xm–1–xm|<∑j=nm1/j, and that sum is not bounded.
Example The metric space Q has Cauchy sequences which do not converge. Any convergent sequence in a metric space is Cauchy (we'll state that formally in a few microseconds). Now take a rational sequence which converges to sqrt(2) in R. It is Cauchy in R, and hence in Q (the metric is the same). But it can't converge in Q since there is nothing for it to converge to. (If it converge to w in Q, the same inequalities would make it converge to w in R. But then since the limits are unique, sqrt(2) would be in Q!)
Example Look at (0,1), the open unit interval in R. The sequence {1/n} is in (0,1). In R it converges to 0, and thus it must be Cauchy, but it can't converge to anything in (0,1) since otherwise it would converge to w>0. But the sequence converges to 0 in R, so (just as before) this is impossible.

A convergent sequence is Cauchy
Use a triangle inequality argument.

If a subsequence of a Cauchy sequence converges, then the original sequence converges.
Use a triangle inequality argument.

A Cauchy sequence is bounded
Indeed, take ε=1. Then the whole "infinite tail" of the sequence after the Nε term is within distance 1 of xNε. Now take a ball big enough to enclose also the finitely many other points left out.

The Direichlet problem, a historical digression
This is a problem in partial differential equations which originated in classical mathematical physics. Attempts to solve this problem in the late 19th centuries led to recognition that "closed and bounded" is not enough to guarantee convergence in many natural "function spaces", specifically function spaces which were used to analyze the Dirichlet problem. This failure or, rather, perhaps, perceived deficiency led to recognition of the importance of compactness, and to the invention of the notions and methods that we are currently discussing.

We went on ...
The lecturer really deliberately tried to investigate the connection between Cauchy and compactness in a rather diffuse non-linear way. This is often how new mathematics develops!. The sequence of ideas went like this:

A metric space is complete if any Cauchy sequence in the metric space must converge. We have just proved that compact metric spaces and any Rk (k finite integer) are complete metric spaces.

The textbook has a much more linearly ordered and careful presentation.

Monday, October 6, 2008

We gave some examples of convergent and divergent sequences. I mentioned the fact that writing "limn→∞xn=x" already seems to incorporate the idea (a theorem, darn it!) that if {xn} converges to both w and v, then w=v.

We then proved (as in the text) that sums of complex-valued convergent sequences are convergent to the sums of the respected limits. Also, if a sequence converges to a non-zero complex number, then "eventually" the sequence is non-zero, and that the sequence obtained by taking reciprocals of the elements of that tail of the original sequence converges, and the limit is the reciprocal of the limit of the original sequence. These techniques are basic and must be part of all mathematicians' subconscious.

I looked at the product of two metric spaces. We put a metric on this product, defined in a way very similar to the Euclidean metric in R2 from d(x,y)=|x–y| in R1. I observed, very briefly, that metrics giving the same topology on the cartesian product are the analogues of the L1 and L2 and L metrics mentioned in the last homework assignment. Then we saw that a sequence in the product converged if and only if the components in each factor (X and Y, respectively) converges. The same is true in any finite product, but with infinitely many factors things get much more complicated. (The last part of the second problem of the Entrance Exam deals with similar obstacles.).The metrics which we considered here (L1 and L2 and L) and which give, in finite dimensions, the same topologies and the same notion of sequence convergence, all turn out to be different in infinite dimensions. This is either distressing and bewildering, or wonderfully enriching!
We will use the observations about products of metric spaces almost always with Rk (where k is a positive integer).

We then moved on to subsequences. I tried to be very careful about defining a subsequence, and admitted my occasional confusion between a sequence (a mapping from the ordered set N to X) and the image of the mapping, a subset of X. Here is a weird example.

Since Q, the rationals, are countable, there is a bijection F:N→Q. Fix any such bijection for this discussion. Now fix any x∈R. I claim there is a subsequence G:N→Q of F so that G converges to x. Why? Well, we must write G as FοI where I:N→N is strictly increasing. I will create I inductively as follows.

  1. What is I(1)? Consider (x,x+1), an interval of positive length. I(1) is a positive integer n so that F(n) is in (x,x+1). This is true since the rationals are dense.
  2. Now we specify I(n) for n>1 inductively. We assume that I(1),I(2),...,I(n–1) have already been defined, with I(1)<I(2)<...<I(n–1). Consider the interval (x,x+(1/n)). There are infinitely many rationals in this interval, else we can easily get a contradiction to the density of the rationals (please: you can and should describe how to get in that case an interval of positive length with no rationals). Since there are infinitely many rationals, there will be a rational q with F(j)=q and j larger than I(n–1). Let I(n) be equal to that j. This inductively defines a subsequence, and clearly, if m≥N, |G(m)–x|<1/n. So the subsequence converges to x.
This proof follows the suggestion of Mr. Skalit made immediately after class.

So the "structure" of subsequences can be quite complicated. I don't think this should be too surprising since there are, after all, uncountably many subsequences.

We stated a result something like this:
Theorem Suppose {xn} is a sequence in a metric space. The following are equivalent:

  1. {xn} converges to x.
  2. All subsequences of {xn} converge to x.
  3. All subsequences of {xn} converge, and they all must converge to the same x.
We will prove this next time. In a general topological space, sequences are not enough (that was the point of the confusing R example discussed earlier). The notion of sequence can be generalized to the idea of nets. Some care needs to be taken.

Thursday, October 2, 2008

Professor Nussbaum was the guest lecturer and did an excellent job. His definition of connected was equivalent to the textbook's definition. For him, X is connected if it can be written as the union of two disjoint non-empty open sets U and V. That the sets should be non-empty is avioding the silly (?) description of X as the disjoint union of X and ∅, of course. If U and V are given, then since each is the complement of the other, both U and V are closed, and the closure of V, say, is closure(V). So we have written X as needed in Rudin's definition of a separation.

He began chapter 3, defining convergence of sequences in metric spaces. He showed that a convergent sequence was bounded, etc. He proved that the product of convergent sequences of complex numbers converged, and its limit was the product of the limits of the factor sequences.

Monday, September 29, 2008

Compactness of k-cells, a proof using bisection
This is the method of bisection. The proof was more or less as in the text, except that the instructor, to avoid being confused, drew a picture for k=2 to guide his statements. The reason for doing everything in the last lecture or so is the following fantastic theorem.

The Heine-Borel Theorem
If E is a subset of Rk, then the following statements are equivalent.

  1. E is closed and bounded.
  2. E is compact.
  3. Every infinite subset of E has a limit point in E.

A counterexample (not really!)
This is a rather simple but still instructive example. Suppose X is any set, and d is the discrete metric. That is, d(x,y) is 1 when x≠y and is 0 when x=y. Every subset of X is an open set always (and every subset is a closed set!). The compact subsets of X are exactly the finite subsets. The diameter of every subset of X is at most 1. So if X is infinite, there are certainly many closed and bounded subsets which are not compact. So the H-B theorem is not true in all metric spaces.

I note that most "infinite-dimensional" situations which arise in analysis (and, indeed, even in certain areas of algebra) naturally many subsets do not satisfy the H-B Theorem. Certainly compactness implies closed and bounded in a metric space, and several textbook homework problems show that the third condition implies compactness in any metric space. But generally the friendly hypothesis of "closed and bounded" will not imply compactness.

Proof of H-B
This was very much like what's in the textbook, with only minor variations introduced by the instructor to ... uhhhh ... challenge the students. Yeah, that's right: challenge the students.

Perfect sets are uncountable
Thanks to Ms. Pritsker and Ms. Hood. Here is her writeup.

The most wonderful perfect set, the Cantor set
I tried to analyzed this set informally. It is an uncountble set. The intervals taken out of it from [0,1] have total length equal to 1, so maybe it should have total length 0. The points in the Cantor set are those whose ternary (base 3) expansions have no 1's. The Cantor set and variations of it are used as the foundation for many disturbing and unintuitive examples in analysis and topology.

It would be nice to tell you that the Cantor set and its relatives are unnatural, etc., except that certain dynamical systems which closely model physical and biological systems have certain types of behavior closely tied to the Cantor set!

Homework due on Thursday, October 9
The third homework assignment is due a week from Thursday.

Out of town
I will be away from Thursday, October 2, to Sunday, October 5.

Thursday, September 25, 2008

The 411 build-a-vocabulary project
Today's word is exegesis. One definition is "Analyzing passages from a document - often the Bible - to understand what it meant to its author and others in the author's culture."

Define of open cover and compact using a finite subcover.

Compactness turns out to be of fundamental importance in many numerical computations. This is not at all clear from the definition and discussion we're about to do. We are beginning at the tail end of a century of struggle with how to handle compactness and related notions, and the "perfection" shown is a bit difficult for a novice to grasp and certainly to understand in a meaningful way.

We verified from the definition that [0,1] is compact.
Here's how Suppose {Gα} is an open cover of [0,1]. Let S={s∈[0,1] : [0,s] has a finite subcover by elements of {Gα} }. Then:

  1. S≠∅ since 0∈S because 0 is in one of the Gα, and that one open set itself is a finite subcover of [0,1].
  2. S is bounded above, since S⊂[0,1] shows that 1 is an upper bound.
  3. If w=sup S, then w∈S. This is true because w is in some Gα, and for some ε>0, (w–ε,w+&epsilon)⊂Gα. If w=sup S, there must be an element of S in (w–ε,w+&epsilon) or else w is not a least upper bound. Let v be that element. Then [0,v] has a finite subcover, and so by including Gα in the subcover also, [0,w] has a finite subcover. So w∈S,
  4. The sup of S must be 1. Just as in the previous section, we see if w=sup S<1, take Gα, and for some ε>0 with (w–ε,w+&epsilon)⊂Gα. If w<1, then including this Gα with the cover tells us that [0,max(1,w+(1/2)ε)] also has a finite cover, which contradicts that w is an upper bound of S.

So here are a bunch of rainclouds, maybe infinitely many, raining on the unit interval. Compactness declares that only a finite number of them are needed for each point of [0,1] to be rained on. (This is a metaphor, darn it!)

We verified from the definition that (0,1) is not compact.
And why this is true As Mr. Baldwin suggested, cover (0,1) by (1/n,1) for each positive integer n. Then the Archimedean property implies this is an open cover (every positive number is greater than some 1/n) and the union of every finite subcover is exactly its "highest" element, and none of the (1/n,1)'s is equal to (0,1).

Balls are enough to check compactness
We need only use open balls instead of open sets in the covers we use to test for compactness.

Suppose K⊂Y⊂X. Then K is compact relative to X if and only if K is compact relative to Y.
I think we used balls here, which might be slightly nicer than the proof of the text. Suppose X=R2 and Y=R1 (considered as a subset of R2 with y∈Y ↔ (y,0)∈X) and with both sets having topologies determined by the usual metric. Notice that the metric in R2 defined by sqrt((a1–a2)2+(b1–b2)2) is just |a1–a2| when we consider points on the horizontal axis. K should be a subset of Y.

Now suppose we have a cover of K by open balls in Y and we know that K is compact in X. The picture to the right is an effort to illustrate the situation. I drew a collection of green intervals. I had to lift them up a little bit ("up" means vertical motion) to show them and not have them overlay and conceal K. If we use the same centers and the same radiuses and create open balls in X, then, since the green intervals cover K, the resulting magenta discs will also cover K. This is because the intersection of each magenta disc exactly equals the green interval which it specified. If we know that K is compact in X, then a finite union of the magenta discs will have K as a subset. The related green intervals will have K as a subset, also, since a point in K is in a magenta disc exactly when it is in the related green interval.
We have proved that if K is X compact then it is Y compact.
Now suppose we have a cover of K by open balls in X and we know that K is compact in Y. So K is in a union of magenta discs. Take one such disc. Its intersection with Y is not necessarily an open ball in Y. It must be an open subset of Y, because each point in the disc is an interior point (we proved that open balls are open!). Therefore each point in the intersection of a magenta disc with Y is contained in a green interval -- that is, an open disc in Y with center in Y and positive radius. This may be a big collection of green sets, but I don't care: the collection of all of these green intervals, for all of the magenta discs, is an open cover of K in Y. Since K is compact in Y, there's a finite subcover by the green intervals. Each green interval comes from one of the magenta discs. So for each green interval in the finite subcover, take the associated magenta disc. The resulting collection of magenta discs is a finite subcover in X of K.
We have proved that if K is X compact then it is Y compact.

Compact subsets of metric spaces are closed.
Proof as in the text. This is a brief and clever proof and the ideas are used elsewhere.

Closed subsets of compact sets are compact.
Proof as in the text.

If F is closed and K is compact, then F∩K is compact.

If {Kα} is a collection of compact subsets of a metric space X such that the intersection of every finite subcollection of {Kα} is nonempty, then ∩Kα is nonempty.
What's written in the hypothesis is called the Finite Intersection Property. One of my colleagues remarked that this result is no more than DeMorgan's Laws. I have always found this confusing. So, with the help of many charitable students, I tried to state the compactness definition carefully and logically. I then wrote the contrapositive. I then used DeMorgan's Laws to translate the statements we obtained. If we consider everything as a subset of one of the given Kα's, we proved the theorem.
This is the proof in the text, where it is given not so histrionically.
Bonus vocabulary word: histrionic Of, or relating to actors or acting; Excessively dramatic or emotional.

If {Kn} is a sequence of nonempty compact sets such that Kn⊃Kn+1 (n=1,2,3,...), then ∩1Kn is not empty.
Proof as in the text. People sometimes say "the Kn's are nested" instead of "Kn⊃Kn+1".

If E is an infinite subset of a compact set K, then R has a limit point in K.

Proof as in the text.

If {In} is a sequence of intervals in R1 such that In⊃In+1 (n=1,2,3,...) then ∩1In is not empty.

Useful and relevant examples

  1. Suppose In is (0,1/n). Then these intervals are non-empty and nested, and their intersection is ∅. But these intervals are open.
  2. Suppose In is [n,∞). Then these intervals are non-empty and nested and closed, but unbounded. Their intersection is ∅. But these intervals are unbounded.

slightly restated 2.38
Make the intervals both closed and bounded. Then the result is true. This phrase has great "resonance" historically.
If In=[an,bn], then an<bm when n and m are positive integers. If α is sup{an :n∈N} and if β is inf{bm: m∈N} then (not totally easy exercise) α<β, and then [α,β] is ∩1In, and certainly the interval is not empty.
Proof as in the text.

Let k be a positive integer. If {In} os a sequence of k-cells such that In⊃In+1 (n=1,2,3,...) then ∩1In is not empty.
We briefly discussed what "k-cells" were.

Here are pictures of 1- and 2- and 3-cells. A k-cell is the Cartesian product in Rk of k closed and bounded intervals, one in each factor.
The proof as in the text.

Monday, September 22, 2008

A metric space has a topology, but not all topologies come from metrics. By "come from" I mean the following process: given a metric, we can define open balls. Given open balls, we can define open sets. The collection of open sets is a topology. One can ask if, given a topology, is there a metric which manufactures this topology using this process. (Thanks to Mr. Baldwin for asking me to clarify what I meant by the informal phrase, "come from" in asking whether all topologies come from metrics.)

This question is more suitably part of Math 441, but we briefly discussed certain desirable (?) properties possessed by metric spaces but not by "random" topological spaces. More generally in the history of topology, a great deal of effort went into discovering which topological spaces occured as a result of a metrics. This is called metrizability and the theorems are intricate.

Uniqueness of limits/Hausdorff
If a topology comes from a metric, then the topology has the following property, which may initially seem rather strange.
Property H Take two distinct points x and y of the space X. Then there are open sets U and V of X with x∈U and y∈V with U∩V=∅.
To see why this is true in a metric space, let q=d(x,y) which must be a positive number since x and y are different. Then put r=(1/2)q, and consider Nr(x) and Nr(y). If z is in both of those sets, then d(z,x)<r and d(z,y)<r, so (triangle inequality and symmetry of d) q=d(x,y)≤d(z,x)+d(z,y)<(1/2)q+(1/2)q=q which is a contradiction since q is positive.

A silly example of a topology which does not have Property H is the following: take X to be a two-element set, say {♣,♦}. The topology on X (which must a set of subsets of X, don't sink into the sea of abstraction here) to consider is {∅,{♣,♦}}. This sure is a topology (called the indiscrete topology, the smallest topology -- the largest topology is called the discrete topology) and it does not have "enough" open sets to separate the two points, ♣ and ♦.
Comment The number of topologies on a finite set is interesting and there's no known exact pattern. Note, oh combinatorists, that most (!!) of them do not have Property H above.

A topology which has Property H is called Hausdorff (the points can be "housed off" from each other [sorry]). This property is already evident in calculus, because we tell people there the following result: if {xn} is a sequence and if limn&rarrow;∞xn=A and limn&rarrow;∞xn=B, then A=B. This "uniqueness of limits" property is exactly a consequence of Property H or Hausdorffness. There are important examples of non-Hausdorff topological spaces which seem less "artificial" than what I gave, but their descriptions are somewhat complicated.

Countable sequence of open sets
A topology which comes from a metric has the following property:

Sequence of neighborhoods
If x∈X, then there is a sequence of open sets, {Un} so that if V is open and x∈V, there must be a positive integer N so that x∈UN⊂V. Actually, we can even ask that these sets be "nested", so that Un+1⊂Un for all n.
Proof A proof is very brief. If d is the metric, take Un to be the ball of radius 1/n centered at x, that is, Un=N1/n(x). If V is as described, there is some r>0 with Nr(a)⊂V. Then the Archimedean property guarantees the existence of N with 0<1/N<r, so that UN⊂V.

This idea will be used again and again in this course. It plays an important part of the "a implies c" proof and you should look at that now if you already haven't. We will use the idea to "create" suitable sequences. But let me show you a somewhat intricate example of a topological space which does not satisfy the preceding result.

As a set, R is "just" the collection of all real sequences. So if x∈R, then x=(x1,x2,...,xn,...) with all of the xn's elements of r. For the purposes of this paragraph, I'll call such an x positive exactly when all of the xn's are positive real numbers. I want to define the box topology on R. I will first begin by defining box neighborhoods.

Suppose x=(x1,x2,...,xn,...) is a point in R and ε=<ε12,...,εn,...) is a positive element of R. Then I will tell you what is in the subset Mε(x) of R. A point y=(y1,y2,...,yn,...) of R is in Mε(x) if and only if, for all positive integers n, |xn–yn|<εn. The sets Mε(x) play a role similar to what the balls Nr(x) do in the case of metric spaces, but there's no metric visible (indeed, there is no metric possible, as we will see!). Now I will tell you what an open set. A subset W of R will be open in the box topology if:
      for all x∈W, there is a positive ε in R so that Mε(x)⊂W.
We should check that the rules for a topology are satisfied. If you wish to do this, you should begin by verifying that Mε(x) is itself open (analogous to verifying that an open ball is open). This is not too difficult, and, indeed, nost of the other verifications are easy. One thing needs to be said, I think. If Mεa(x) and Mεb(x) are two of the defining neighborhoods, and if we put εc;=(min(εa1b1),min(εa2b2),...,min(εanbn),...) (take minimums in each coordinate) then M&epsilon3(x)⊂Mεa(x)∩Mεb(x). This result is needed to prove that intersections of box open sets are box open.

Now suppose that x∈R. What if there were a qualifying sequence of box open sets {Un} behaving nicely as in the Sequence of neighborhoods proposition above? Then there must be (definition of box open) a sequence of positive εn's so that x∈M&epsilonn(x)⊂Un. I will create a specific Mη(x) which is not contained in any of the M&epsilonn(x)'s.

Let me begin by looking at a specific example, because that will help us understand the general case. Take Un to be M(1/n,1/n,...,1/n,...)(0). So this is a neighborhood of 0 whose "polyradius" is 1/n in each coordinate or direction. Certainly, there is no point other than 0 in all of the Un's, so they seem to "shrink down" to 0. As a response to this choice of a sequence of Un's, I ask you to consider V=M(1,1/4,...,1/n2,...)(0). I claim that none of the Un's can be included inside V. Why? Well, let's look at a point pn in Un. Define pn to be (2/(3n),2/(3n),...,2/(3n),...) so it is 2/(3n) in each coordinate. This is certainly inside Un since 2/(3n)<1/n. But it is not in V, because 1/n2 is eventually less than 2/(3n).

Now let's try something similar in general. You "challenge" me with a list of Mεn(0)'s.
My Mη(0) will be constructed so that its sequence of coordinates →0 faster (eventually!) than any of the sequences of coordinates of the εn's. Here is one possible recipe:
     Take ηj=(2/3)min(1/j,ε1j2j,...,εjj)
Then what do I know? I know that &etaj is much smaller than the jth coordinates of ε1, ε2, ..., εj. The element of R defined by two-thirds of &epsilonn is in Mεn(0), surely (I just mean, multiply the components of &epsilonn each by 2/3). But it is not in V. Why? Because for j>n, ηj is less than (2/3)εnj. This "tail" restriction prevents the element of Un from being in V.

What's going on?
Given a sequence of sequences of positive numbers, we have created a sequence whose limit is 0 and which approaches 0 eventually faster than any of the sequences. The "eventually" is that it is smaller than the given sequences at a variable starting place, of course. This is again the diagonal process applied in a more complicated way.

Please note that in class I used a much simpler prescription to create η which won't necessarily prove what is needed. I am sorry. I also owe emphatic thanks to Ms. Slusky whose persistant inquiries and messages made me think a bit more about this. It is more subtle than what I hurriedly did in class.

First countable
A topology which does satisfy the Sequence of neighborhoods proposition is called first countable. I showed that a topology obtained from a metric space is first countable, and gave an example of a topological space which is not first countable. There are simpler examples but the example given is also Hausdorff, and the underlying set, R is neither artificial nor unimportant, but arises naturally in probabilistic reasoning (select a random sequence of real numbers). The box topology is not used in probability because it has other defects besides lack of first countability. In a first countable space, sequential reasoning is adequate to determine what is and is not an open set.

Metric spaces are Hausdorff and first countable, and I likely never refer to these terms officially again in this course, but I will use the properties they define frequently.

More definitions
Now for the actual progress of the course. We return to metric spaces,

Interior point; interior of a set
We defined interior point and the interior of a set. We prove that the interior of a set is equal to the union of all the open sets which are a subset of the set. We investigated some examples.

Closed set
A closed set is the complement of an open set. Some sets are neither open nor closed. We investigated some examples. The closure of a set is the intersection of the closed sets containing the set. We characterized this using the idea of limit points, and considered (rapidly!) some examples. A closed set is one which contains its limit points.

Perfect set
A perfect set is a set which is its own limit points. Ms. Pritsker, with the help of Ms. Hood, volunteered to prove, some time, that a perfect set in a metric space is uncountable. Thanks to them in advance.

I did not define "dense" and will need to begin next time with that definition. Sigh.

Thursday, September 18, 2008

A few more remarks about cardinality
I reviewed what we "knew" so far. Perhaps the most interesting additional remark I made concerned the Schroeder-Bernstein Theorem which says, roughly, if }A|≤|B| and |B|≤|A|, then A~B. More precisely, it states that if there is an injection (1-1 mapping) from A and B and if there is an injection (1-1 mapping) from B to A, then A~B (A and B have a bijection, so they are "the same size"). I will not prove this here -- the proof is "elementary" but it is not part of Math 411. If you are curious, take a course or read.

The real numbers are uncountable
Here is what I said. This is, more or less, Cantor's second (!) proof of this assertion (more later). Suppose R were a countable set. Then, since subsets of countable sets are countable, the interval [0,1] would be countable. Now supponse [0,1]~N, the positive integers. Then there would be an "enumeration", that is, a way of listing the elements of [0,1] according to their image under a hypothetical bijection with N. So let us enumerate the elements of [0,1], listing each one according to its decimal expansion address.

First element <---> .a11a12a13a14...
Second element <---> .a21a22a23a24...
Third element <---> .a31a32a33a34...
Fourth element <---> .a41a42a43a44...
Now we define a number b in [0,1] by giving its decimal expansion address:
where bj=7 if ajj is not 7 and ajj=4 if j=1. This is silly but effective, really. Notice that b can't be equal to any of the "elements" listed in the enumeration because the decimal expansion address is different. So we are done.

Criticism of this Cantorial proof
Well, the problem is that the decimal "address" is not unique. What do I mean? By analogy, if we talked about a building at the intersection of Fourth Avenue and Avenue A, it is logically possible that the building would have two distinct addresses. It could be called, say, 47 Avenue A and 222 Fourth Avenue. Certainly that occurs with decimals. For example, I know (?) that .379999999...(9's repeating) is a decimal address for a number which also has the decimal adress .38000000...(0's repeating). Well, hw many numbers have that problem? The problems are those numbers with decimal expansion "ending" in an infinite string of 9's. Well, golly, how many such problematic numbers are there? For any finite string of integers, .c1c2c3...cL I could end with 9's repeated or subtract 1 fron cL. (This is not exactly correct, since cN could be 0 but I'm getting tired.) So this countable set of reals has maybe two addresses. Throw them out and apply the Cantor proof (the "Cantor diagonal process"). The result will be an unenumerated real number. So the proof works. Let's follow Professor Rudin's text, though. This will mean using the diagonal process you have already seen at least two more times, and maybe even more. This is not a bad thing, since the idea is really inspired, first-class, etc.

Really, a set which is not countable
Let's consider the sequences whose values are 0 or 1. That is, such a sequence is a function, f:N (the positive integers)→{0,1}. Or, if you are a traditionalist, it is {an}n in N where the an's are 0 or 1. What if this set (let us call it, temporarily, S) is countable? We enumerate the sequences in S. That is, S consists of f1, f2, f3, f4, ... and we "diagonalize" to create a sequence which is not enumerated. So g(n) is 1-fn(n). I wrote this equation and knew this was sort of being exhibitionistic -- I just wanted to show you I could. Sign. What the heck is g? At the positive integer n, is is 1-fn(n). So (work it out, I have, darn it!) this "switches" the values. That is, if fn(n) is 1, then g(n) is 0, and if fn(n) is 0, g(n) is 1. So g can't be any function already counted. The set S cannot be enumerated, and it is not countable.

Hash [huh?]
Let me now try to create an injective function I:S→R. I will be wrong, but, I will try! First, as preface, I will assert the following wonderful fact:
If n is a positive integer and a is non-negative, then ∑j=1nrj=(r–rn+1)/(1–r).
Now you may know this as the partial sum of a geometric series, but I know it in Math 411 as a fact which can be confirmed with mathematical induction if r≠1. Well, if I know this fact, then (with a≥0 and 0≤r<1) any such sum is positive and certainly less than r/(1–r). for example, if r=1/2, the sums will all be less than 1.

Let's try the following to create an f. If A={an} is a sequence of 0's and 1's, define I(A)=sup{∑n=1Nan/2n : N is a positive integer}.

There's a bunch of things to check. First, does the set indicated above have an upper bound? The key observation is that
So the set involved in I(A)'s definition is bounded above, and therefore it has a sup. Unfortunately, several students pointed out that the Emperor's clothing is indeed lacking. I is not 1-1. It is, after all, just the sequence {an} encoded as the binary expansion of a number. And just as decimals have problems, so do these expansions. For example, the sequence {1,0,0,0,...(all 0's)} gets mapped to a set whose only element is 1/2. But the sequence {0,1,1,1,...(all 1's)}: well, we need the sup of ∑n=2N1/2n={1/2}–{1/2N). Since we proved that 2N is unbounded, these sums also have sup equal to 1/2. The mapping I is not injective.

How to fix it
Computer science sometimes calls functions like I by the name, hash function (this is part of the special Math 411 Vocabulary Building Project). The hash function classifies elements of a set. Here the set to be "classified" are the collection of 0-1 sequences, and we will classify them by real numbers in [0,1]. When its values of the hash function are the same for different inputs, the result is called a collision. Can we redefine I to avoid collisions?

One suggestion (from Mr. Kowalik?) was to replace 2 by 3 in the definition of I. This will work. In fact, we will use something similar next week in a slightly different context. I wanted to use a weirder idea, because it leads to a wonderful fact mentioned later. So my "suggestion" was to replace 2n by 2n!. This means that we are filing the sequences into bins (?) of [0,1] which have widths which shrink much more quickly. Let's now consider two different sequences, say A={an} and B={bn}. Let's assume that N is the first time that they disagree. So an=bn for n<N and, to be specific, aN=0 and bN=1. I claim that I, as modified with n! instead of n for the exponent, will have I(A)<I(B): they won't be equal. If I verify this, then I is injective, and, since S is uncountable, we will have verified that R is uncountable. Wow!

Let's make an I(A) as large as possible and an I(B) as small as possible, and compare them. The initial segments appearing in the sum are the same, and if I show what follows, then we're done:
Now consider the sum. Its first term is 1/2(N+1)!=1/(2N!)N+1 (repeated exponents are subtle!). What about the ratio between successive terms? Well, the factorials make what's written above not a geometric series. So let me overestimate what is above by a geometric series with initial term 1/(2N!)N+1 and ratio between successive terms equal to 2(N+2)!–(N+1)! (this is just the ratio between the first and second terms -- all the other ratios are much less because the exponents are even larger). The exponent is at least 4. So the series given above is less than a geometric series whose first term is 1/2(N+1)! and whose ratio is 1/4. The sum of that series is [1/(2(N+1)!]/(1–{1/4}) which is 1/(3·2(N+1)!–2). Now let me ask about this inequality:
     When N=1, this is 2<3·20 which is true.
     When N=2, this is 22<3·24 or 4<48 which is true.
     When N=3, this is 26<3·222 which is very true (heh, heh: 64<125,582,912: silly).
The desired inequality is always true, and can be "left to the student" (factorials grow very fast!). So we are done, and we know the reals are not countable. We will see other proofs of this fact, but now let's use it.

Transcendentals ...
Mr. Ratner discussed problems 2 and 3 in chapter 2. He defined algebraic numbers and showed that they were countable, and therefore concluded that the transcendentals, the "other" real numbers, were uncountable. Here is his writeup.
Note For the algebraists among you, please realize that the algebraic numbers are a field. This is not totally trivial, but it is algebra, so I can't prove it, and I don't need to in Math 411.

Opinion and history
I said as I believe, that this sequence of silly trivial observations is truly remarkable. One consequence is that most real numbers, if the word "most" is used in almost any sense, are transcendentals. So a random real number should be transcendental. It is rather difficult to make sense of the word random in an infinite setting, which is one of the real difficulties of probability.

Cantor's proof of the existence of transcendental numbers as described above is the only one I knew until fairly recently, but apparently it was his second proof. Earlier in his life, he published a proof which can be described as constructive. A discussion of this proof is in the article Georg Cantor and Transcendental Numbers by Robert Gray, in The American Mathematical Monthly, Vol. 101, No. 9 (Nov., 1994), pp. 819-832. If you are at a Rutgers computer, you should be able to see a copy of this paper using this link.

Yeah, there are lots of them; now show me one!
I know because people have told me that the special numbers e and π are transcendental. The original proofs of these facts are difficult and long, and even now, a century later, I do not think there are any really easy proofs. It would be nice, since most real numbers are transcendental, to give at least one explicit example of such a number. Here:
The transcendentality of this number is a result of Liouville, whose name is also attached to other results in math and physics. The foundation of the rather brief proof is the Mean Value Theorem of one variable calculus, and the strange statement that algebraic numbers which are not rational can't be approximated very well by rationals in a suitable sense. An exposition is here.

We must go on!
The central concerns of the course will frequently be stated in terms using topology. So we should study a bit of topology. We have a whole course, Math 441, devoted to this. I've taught it. Let me outline what my first lecture in that course was about.

How to begin Math 441
My assumption was that not all students were knowledgeable about advanced math and proofs, but that there would be some recall of calculus. I "remembered" some definitions and introduced a definition new to many students.


Most students vaguely remembered the first two definitions. I tried to give some "intuition" for the third by remarking that due to the difficulties of measurement, we rarely would observe the exact value of a w, but usually only had some fuzzy knowledge about it (up to an error of ρ). Then I stated a result.

Theorem Suppose f:R→R is a function. The following logical statements about this function are equivalent.

  1. If {xn} is any convergent sequence with limn→∞xn=L, then the sequence {f(xn)} also converges, and limn→∞f(xn)=f(L).
  2. f is continuous at every point.
  3. If U is any open subset of R, then f–1(U) is an open subset of R.
All this is a great deal of abstraction, and I tried to explain why these statements were logically equivalent. That took more than an hour in Math 411. Here, a request was given to verify one implication. I think it was b implies c. Let's try this.

Proof of b implies c
I will try to verify c, and somewhere I will use b. So suppose U is an open subset of R, and w is in U. Since U is open, I know there is ρ>0 so that (w–ρ,w+ρ)⊂U. This interval translates to the statement: all real y's which satisfy |y–w|<ρ (interval statement and inequality statement are logically equivalent). Suppose w=f(a). Then the inequality becomes |y–f(a)|<ρ. Now use b which declares "there exists" (magic!) δ>0 so that if |x–a|<δ then |f(x)–f(a)|<ε. But think about this: it means in turn that the set f–1(U) includes the numbers x which satisfy |x–a|<δ. But that means (inequality/interval) (a–δ,a+δ)⊂f–1(U). Think this through: we have produced an open interval contianing a inside f–1(U), and we can do this for any a inside f–1(U). Therefore f–1(U) is open.

There were no other requests until later that afternoon brave Mr. Leven and brave Mr. Kiria came to my office, risking GBH (in English mysteries, GBH is "grievous bodily harm" but I find to my horror that GBH is also a text-messaging abbreviation for "great big hug"). They worked through what I think is a more difficult proof involving sequences. Let me present their proof here.

Proof of a implies c
I need to convince you that the sequential statement implies the open/inverse open statement. In fact, here we will proceed by proving the contrapositive: if the open/inverse open statement is false, then the sequential statement is false.

Let's suppose I know an open subset U of R so that f–1(U) is not open. Well, this means there's at least one w in U with f(a)=w and one ρ>0 so that (w–ρ,w+ρ)⊂U, so that there is no τ>0 with (a–τ,a+τ) inside f–1(U). Let's use this complicated statement to create a sequence. If n is a positive integer, take τ to be 1/n. The interval (a–1/n,a+1/n) is not a subset of f–1(U), so there must be a number xn in this interval with xn not in f–1(U). That is, |xn–a|<1/n and f(xn) is not in U. Since f(xn) is not in U, it also can't be in the interval (w–ρ,w+ρ) because that interval is a subset of U. Therefore we know that |f(xn)–f(w)|≥ρ.

Putting this all together, we have created a sequence {xn} so that |xn–a|<1/n. Therefore (Archimedean property) this sequence converges to a. But w=f(a), and |f(xn)–f(a)|≥ρ where ρ is some positive number. The sequence {f(xn)} can't converge to f(a)! And a must be false! We have verified if c is false, then a is false. Therefore a implies c.

YOU TOO can try to prove one of the implications guaranteed by the theorem above. Tell me about it, or ask me about it!

So what people then did ...
Historically, people decided that statement c was a neat way of studying ideas like continuity. It seemed to change inequalities and complicated implications into "simple set theory" (well, the packaging was good, but the difficulties are just hidden, not gone!). So a whole industry was created to understand this sort of approach.

A metric
I defined metric as in the text, commenting that most difficult rule to verify was usually the triangle inequality. I gave two examples. The most important example for Math 411 was the Euclidean metric (square root of the sum of the squares of the differences in the coordinates). The triangle inequality then is a consequence of the Cauchy-Schwarz inequality. The silliest example of a metric is the so-called discrete metric, where d(x,y) is 1 if x≠y and 0 if x=y. Note that d(x,y) will play the role of |x–y| in arguments similar to what we just went through.

More definitions
An open ball or open neighborhood was defined as in the text: Nr(p) is the collection of x's in X with d(x,p)<r. This plays the part of (p–r,p+r). The open balls in Rn are "clear". In the discrete metric, an open ball of radius<1 is just 1 point. If the radius is ≥1, it is everything. An open set is one which contains an open ball centered at each point in the set. The radius, r, of the ball will usually depend on the point selected! The open sets for the discrete metric are all subsets of the set!

So use these properties to define ...
A topology on a set X is a set of subsets of X, let's call it Τ, with the following properties:

  1. ∅ and X are in Τ.
  2. Finite intersections of elements in Τ are in Τ.
  3. Any unions of elements of Τ are in Τ.

I'll give some more examples next time, and go through more of chapter 2.

Monday, September 15, 2008

Aim(s) of the course
To get students comfortable with inequalities and with abstraction, and to encourage a certain self-critical quality when considering one's own proofs (or instructor-critical if nothing else!).

I gave a good proof of the thing I messed up last time.

We discussed the complex numbers, C, as R[i]/(i2+1) and as R2 with vector addition and a weird kind of multiplication and as 2 by 2 real matrices of a strange kind (complex conjugation is realized as transpose of matrices in this guise -- I was, as usual, incorrect!).

I verified a few of the elementary properties of complex numbers, as in the text. I then started to prove the Cauchy-Schwarz inequality. The method I used might have looked a bit different, but was actually the same, as what was in the text. I considered the function f(t)=||A+tB||2 where A and B are vectors in Cn. We knew that f(t)≥0, so when I "expanded" this function using bilinearity, the discriminant of the resulting quadratic was non-positive. This is almost the Cauchy-Schwarz inequality, and I tried to suggest how it would verify the text's version of Cauchy-Schwarz by multiplying one of vectors by the scalar e for suitable choice of θ (to make the appropriate dot product real and positive).

I discussed enough of Rn to be sure that we could recognize it as a metric space in a few days. In particular, I got versions of Cauchy-Schwarz and the triangle inequality.

Sizes of sets (cardinality)
It will be useful to have this language as we discuss certain examples. So I defined (with help!) the following terms:

  1. A~B means
    A and B have the same cardinality, same size: there is a bijection between them. Then A~A (the identity as a bijection); if A~B then B~A (the inverse of a bijection is a bijection); if A~B and B~C, then A~C (the composition of bijections is a bijection). We will get into serious paradoxes if we consider ~ as an equivalence relation on all sets, however.
  2. A is finite if either A empty or A~{1,2,...,n} for some positive integer n.
    For a fixed non-empty finite set, the integer n is unique (takes some proving!).
  3. A is countably infinite if A~N, where N is the set of all positive integers. Examples of countably infinite sets include N and Z (all integers) and Q, the rationals. There are various sources for an explicit bijection of Q with N (an enumeration) such as Proofs from THE BOOK by Martin Aigner, Günter M. Ziegler, and K.H. Hofmann. However, this follows as a consequence of the fact that the countable union of countably infinite sets is countably infinite. This "fact" uses a Cantor diagonal argument.
  4. No "biggest" set exists! Given a set A, there is a set B so that no surjection (onto mapping) exists, and therefore there is no bijection, so A and B are not the same size. This is proved using a version of the barber paradox. Here the following notation is customarily used: if A is a set, then 2A is the set of all subsets of A. Why this notation? Well, 2={0,1} (!) and the notation 2A indicates the mappings from A to {0,1}. A subset T of A corresponds to the function which is 1 on elements of T and 0 otherwise (the indicator or characteristic function of T).
    Proof: Suppose f:A→2A is a candidate onto mapping. Define T to be the subset of A defined by {x is in A if x is NOT in f(x)}. If T=f(x) for some x, then either x is in T, so x is in f(x) which means x is in T but then (definition of T) x is not in f(x) which is x, or x is not in T, so x is not in T=f(x), so x satisfies the determining property of T and x is in T. Both assumptions lead to contradictions, so no such f exists.
    Of course this now means there is no"largest" set, since we can keep taking "subset of ... subset of ...". The universe of sets can be difficult to understand.
  5. A set is countable if it is either finite or countably infinite.
Next time I hope to prove that the real numbers is not countable. Such sets will be called uncountable. Since from what we already know, the set of all subsets of N is not countable, this brings up the Continuum Hypothesis, which is whether the real numbers, an uncountable set, necessarily must be the same size/cardinality/~ as the set of all subsets of N since both of these sets are "larger than" N. This is a very famous basic question of set theory. Mr. Ratner volunteered to present problems 2 and 3 of chapter 2, with his preparation aided by Ms. Pritsker.

Thursday, September 11, 2008

Again the class began by declaring that the real numbers, R, were a complete ordered field, and, up to field isomorphism, the only such. The construction of R from Q had been incompetently discussed in earlier course meetings.

So non-empty subsets of R which have upper bounds will always have least upper bounds, called sups. We verifeid a few standard simple consequences of completeness.

The Archimedian Property
If x>0 and y are real numbers, there is a positive integer N so that Nx>y.
This was proved last time. To me there is a geometric contect, that if we lay out a "grating" of width x on the real line, we will always "trap" y inside the grating. If you believe that the Archimedean property is somehow intuitive then probably you have not met the long line which locally looks just like R but is much, much longer. It is irritatingly unintuitive.

The Archimedean property says that the real numbers can't contain infinities. Take reciprocals, and we see that infinitesimals can't exist either (a pity, because many analysts, me included, frequently think about proofs with them).

If x>0, then there is a positive integer N so that 0<1/N<y.

I needed a simple inequality to get a multiplicative version of the Archimedean property. In some textbooks this is called Bernoulli's Inequality. Here it is:
If x>1 and n≥2, then xn>1+n(x–1).
This can be proved directly using mathematical induction (verify for n=2; assume for n, multiply both sides by x and juggle some algebra to get the statement for n+1). Or if you know the Binomial Theorem, look at the first two terms of the expansion of xn=(1+[x–1])n.

A sort of multiplicative Archimedean property
If x>1 and y are real numbers, then there is a positive integer N so that xN>y.
This is true because we can select N so that 1+(N–1)x>y, and then use the previous lemma.

These observations will be used again and again.

Then I turned to more ambitious verifications. I said there were three statements which the proved directly from completeness. Here is what they are, in order (to me!) of increasing difficulty and here is what we did (and will do). All of these statements could be quite easily proved once the machinery of one variable calculus has been developed, so these direct proofs, while they are all elementary, all involve some intricate contortions of logic.

  1. If n is a positive integer, then there always are nth roots of positive numbers. That is, if y>0, we solve the equation xn=y with x>0.
    This is the easiest, and the proof is laid out in the book, and the instructor will read the book to the class.
  2. If b is a positive real, we can define bx for any real number x is an algebraically satisfying way, extending and consistent with our previous uses of the exponentiation notation.
    This is more intricate. It is problem 6 in chapter 1. Only an outline is presented there, and a number of irritating, mostly but not all (to me) easy details need to be verified. Ms. Slusky will present this, with her preparation aided by Ms. Sergel.
  3. Logs exist.
    This may be even more intricate. It is problem 7, and I asked you to write a solution and hand it in at Monday's class.

Roots exist Given y>0 real and a positive integer n, there is a unique x>0 so that xn=y.
Proof I won't write out the details, since I basically copied what was in the text except that I tried to show how a proof of this result could be discovered. So we consider the set of real numbers, S={x: xn<y}. The logical outline of the proof follows:

  1. S is not empty.
  2. S is bounded above.
Then we know that sup S exists, and let's call this number, s.
  1. sn<y leads to a contradiction.
  2. sn>y leads to a contradiction.
Then the Law of the Excluded Middle (trichotomy) tells us sn=y.

I proved 1 as in the text: take y/(1+y) which is a positive number less than 1 and less than y. Integer powers of numbers between 0 and 1 "shrink" so this provides an element of S.

I proved 2 as in the text: take 1+y, larger than both 1 and y. Integer powers increase, so 1+y is not in S, and, because x1<x2 implies (x1)n<(x2)n we know that no number bigger than 1+y is in S, and therefore S is bounded above. Notice that the observation given also implies that nth roots, if they exist, must be unique.

Then I tried 3. There is a complete proof, awesomely well-arranged, in the text, so I tried to understand how that proof might have been invented. Well, if sn<y, we can try to kick "up" or increase s and get another number whose nth power is still less than y. So let's look at s+h with h some "small" positive number to be determined. Well, notice that (s+h)n≤sn+STUFF. We'd like this to still be <y. So we need STUFF<y–sn, and y–sn is some random positive number. So how can we choose h? This is the origin of Rudin's specification. You could try to finish this yourself, please, with the book and notes closed. Here is the case n=3.

STUFF=h(3x2+3xh+h2)≤ h[max(3,3,1)x2] if h≤ x. So to get this less than y–x3, take h positive which is also less than (y–x3)/[max(3,3,1)x2], and b is done for n=3.

4 is proved similarly.

I made a very bad mistake
What I did to prove 3 in class was wrong and I was too excited and went too darn fast to notice. I am sorry. So what did I do and how did my spurious argument go? I had sn<y and wanted h>0 so that (s+h)n<y. Then somehow (stupidity!!!) I used the true observation that sn<sn+hn<(s+h)n combined with selesting h so that sn+hn<y to conclude that I had found a satisfactory s+h (one with (s+h)n<y). This is an inference which is NOT valid. So: a<b<c and b<d does not necessarily imply that c<d (find some numbers, if nothing else, and use that to convince yourself!).

I am sorry. The comments above for n=3 show how I should or could have correctly ended the proof.

Problem 6
The existence of real powers of positive numbers was nicely discussed by Ms. Slusky. Here is a pdf she supplied for the class. I thank her for her work.

Monday, September 4, 2008

The lecturer did not do very well at all. He/I attempted to describe certain aspects of the formal verification that Dedekind cuts are a field. I thought this would take 12 to 15 minutes. Hah. I spent more than an hour, and got almost nowhere. "It's all in the book." If that is correct, what is the purpose of the class? Don't ask. The remarkable and wonderful fact that the rational numbers are Archimedean contributes importantly to the proof that some of the field axioms are valid for the real numbers, considered as a set of cuts.

I did prove one easy consequence of completeness, the Archimedean property, and the further consequence that there are no positive reals smaller than 1/n where n runs through all positive integers. The Archimedean proof would be an offense to constructivists, and the fact about 1/n would be false to folks who like infinitesimals (that probably includes most analysts!).

Maybe things will go better next time. It can't be worse. We will be treated to Ms. Slusky, coached by Ms. Sergel, giving a beautiful presentation of problem #6 on irrational powers. It's gotta be better than ... well, enough ...

My reasons for volunteering early

Thursday, September 4, 2008

The word gloss has a number of meanings. Some of them include the following:
  1. an explanation or translation, by means of a marginal or interlinear note, of a technical or unusual expression in a manuscript text.
  2. a series of verbal interpretations of a text.
  3. a glossary.
  4. an artfully misleading interpretation.
I think what's written here is meant to be 1 or 2 above (I hope not 4!). I don't know how much time I'll be able to put in, since I am teaching an extra course this semester. But let me try.

We investigated the statement

R is a complete ordered field

A field is a set with two "operations", addition and multiplication, each a functon from pairs of elements of the set to the set. Each operation is commutative and associative, and each has an identity. Inverses also exist (except that 0 doesn't have a multiplicative inverse otherwise things would be very silly!). And there's a distributive law, relating multiplication and addition.

With no further elaboration, there are many examples, highly varied, ranging from finite fields to "big" examples such as fields of rational functions. Field theory is rich enough to totally support such intellectual enterprises as linear algebra and investigating the relations between fields is an interesting and worthwhile part of algebra.

We rapidly worked through a few simple consequences of the field axioms (pp.6-7) such as –(–x)=x and 0x=0.

An ordered set in the text is a set with a relation, <, which is transitive (x<y and y<z implies x<z) and has trichotomy (exactly one of these statements is true: x<y, x=y, y<z). An ordered field is an ordered set which further obeys these rules: if x<y then x+z<y+z, and x>0, y>0 implies xy>0.

In other texts an ordered field is a field which contains a subset P (of "positive" elements) so that one of these is true: x=0 or x is in P or –x is in P; and P is closed under addition and multiplication. We verified that our text's defintion agreed with this one with x<y defined as y–x is in P. Also we checked a few simple consequences of the ordered field rules (p.8). It is true that an ordered field has characteristic 0. I didn't state this, but if you know what characteristic means, you can probably verify it.

An upper bound for a subset W of an ordered set is an element t so that for all w in W, either w<t or w=t (together written as w≤t). A least upper bound of a set W is an upper bound which is ≤ any other upper bound. Trichotomy implies that a least upper bound, if it exists, must be unique. Such an element will be called sup W. The dual notions are lower bound and inf, the greatest lower bound.

Finite ordered sets have sups (they are actually maxes and mins). For a more relevant and interesting example we used Q and its subsets.

The rational numbers, Q
The text assumes known the ordered field of rational numbers, Q, together with its properties and its representation as quotients of integers. This information can be deduced from more primitive notions (see some of the references) but we begin here in 411.

Q itself does not have an inf or a sup.

The important definition that I forgot (at first) in class is of the word complete. An ordered set is complete if every non-empty subset with an upper bound has a sup. The word "non-empty" is important, since the empty subset has many upper bounds but may not have a sup (check this for the empty subset of Q).

Q's major defect from the analysis point of view is that it is not complete. The usual discussion involves "sqrt(2)". The reason for the quotes is
There is no rational number whose square is 2.
You should know how to prove this. Does the proof work for 3? Does it work for 4? Why or why not?

Notice that this statement does not necessarily imply that Q is not complete. A similar statement ("no rational number has square –1") is true for both Q and R, although R is complete. The distressing thing about the statement is that it shows the need for completeness if you want nice calculus. For example, look at the graph of y=x2–2 in the "rational plane", Q2. This graph violates the conclusion of the Intermediate Value Theorem!

Suppose C is the collection of rational numbers whose square is less than 2 or which are negative. This is a silly looking set, an infinite half-line. The reasons for considering this particular set will become clear later. Then:

  1. C is bounded above (by, say, 3).
  2. If v is a rational upper bound of C, there is a rational upper bound w with w<v.
  3. If a rational number v is in C, there is a rational number w in C with v<w.
It is not completely obvious how to prove this. That is, how to verify that such w's exist given v's. You can't reference sqrt(2) in your desire to get the v's since there is no sqrt(2) in Q! In fact, there are procedures for getting better rational approximations to sqrt(2) given a rational approximation. That is, given v, for finding an appropriate rational w with |w2–2|<|v2–2|. The procedures are not totally obvious. I know two of them. One is Newton's Method (input a rational, get out a rational) and one follows from the Continued Fraction expansion of sqrt(2). Both need some sophistication, but the result can be used. One is shown in example 1.1 (p.2) of the text but of course the source of the formula is not discussed. So once you know the complicated statement above, you would agree that C has no sup (in Q!) and therefore Q is not complete.

A major achievement, possible to do in a number of ways, is to show the following:

  1. There is a complete ordered field. I'll discuss this but not verify all the details since that would probably take 3 or 4 class meetings!
  2. All complete ordered fields are "order field isomorphic". That is, given any two such, there is a bijection F which "preserves" addition and multiplication and order: F(x+y)=F(x)+F(y) and F(xy)=F(x)F(y) and F(x)<F(y) whenever x<y.

This is a really nice fact. I will not go through the proof in detail, since we won't need any of these techniques later in this course. It is certainly a part of "mathematical literacy" to know something about these ideas, and similar ideas are used in other mathematical fields. I hope to show you enough so that you can see how clever it is, and also so that you can see the difficulties.

"Construction" of R from Q using Dedekind cuts
Our complete ordered field, which we will call the real numbers, R, will be a collection of subsets of Q. This is a rather sophisticated notion. The subsets are called Dedekind cuts or just cuts. So: a subset C of Q is a cut if

  1. C is neither the empty set nor all of Q.
  2. If w is in C and v<w, then v is in C.
  3. If w is in C, then there is v in C with w<w.
The second and third statements need some thought. They essentially declare that C will be a left half-line which does not contain its right endpoint. The reason why this is not totally correct is that the set of endpoints includes not only things like 1 but also things like "sqrt(2)" which is not in Q as we saw. So the example of the C earlier is a Dedekind cut not corresponding to an element of Q.

The text describes this whole setup, with proofs, in less than 5 pages of the Appendix to Chapter 1. A more leisurely (25 pages!) discussion is available in Spivak's Calculus book which you can borrow from me.

I won't go through all details because that would probably take 3 or 4 lectures. I will try to do enough so that you can appreciate the cleverness and also understand the intricacy of certain verifications. I won't need the techniques of these proofs later in this course, but I will remark that knowing about the setup is certainly part of everyone's mathematical literacy, and also analagous constructions are used in other fields with great success.

Cuts and <
If C1 and C2 are cuts, then we define C1<C2 to mean: C1 is a proper subset of C2. (Cute!)

Completeness of the set of cuts
It turns out with all of these definitions, completeness is easy. Since this is what is usually quite difficult, maybe the payoff is right here, upfront. So what do we do? Suppose {Ca} for a in A is some non-empty set of cuts which is bounded above by the cut D. This means that each Ca is a subset of D. Let C be the union of all the cuts Ca. I claim that D is the sup of the set of Ca's. We need to verify:

  1. C is a cut. This is usually the most irritating part, as you will see. Here it is not difficult. So: since D is a cut, D is not all of Q, and therefore the union of the Ca's is a subset of D and is not all of Q. It is also not empty, since the set A is not empty by assumption. Also, if w is in C then w is in some Ca. So all v's less than w are in that Ca and these v's must be in D. Similarly, if w is in C and thus in some Ca we can find v larger than w in that Ca and thus in C.
  2. C is an upper bound of the Ca's. True since each Ca is a subset of C, so each is either < or = to C.
  3. C is a least upper bound. If E is another upper bound, then for each a, Ca≤E, so the union of the Ca's is a subset of E. And that's exactly C≤E.
Isn't this neat!?

This method makes verifying completeness easy. Some of the extensive details involving the algebra are difficult. Let me just discuss addition.

Addition of cuts
If C and C' are cuts, we will define a subset of Q to be called C+C'. So z is in C+C' if z=x+y for some x in C and y in C'. We now need to verify that C+C' is a cut, we need to check commutativity and associativity, identify and check an additive identity, and look for additive inverses. I'll do some of this next time.

Maintained by and last modified 9/5/2008.