Comments on the first exam in Math 411, fall 2008

Grades ranged from 48 to 95. Letter grades and other specific comments on the the inside of the first page of the exam.

Problem 1 (13 points)
Of course this is the famous (?) Squeeze Theorem which is used several times in elementary calculus.

Problem 2 (13 points)
Some definition of connected should be used. Imprecise use of "pieces" or "parts" is not sufficient. I also searched for explicit use of the hypothesis that P∩Q is not empty, since the result desired (P∪Q connected) is not generally true without it.

Problem 3 (15 points)
a) (11 points) Here I looked closely for a specific use of invalid logic. Consider these statements:

A ∀r>0, Nr(x)∪(A∩B) is not empty.

B At least one of the following statements is true:
     ∀r>0, Nr(x)∪A is not empty.
     ∀r>0, Nr(x)∪B is not empty.

Some students claimed that A and B are equivalent. This is true because of the nested nature of Nr(x) as a "function" of r. That is, if r1<r2, then Nr1(x) is a subset of Nr2(x). Without noting that or some equivalent statement, the equivalence desired is not generally valid. Why? Think carefully about quantifiers. Here is an example.

Suppose we define the symbol Sr(x) for r>0 to mean the following:
     If r∈[1/(n+1),1/n) and n is even then Sr(x)={0}.
     If r∈[1/(n+1),1/n) and n is odd then Sr(x)={1}.
Notice that if A={0} and B={1}, then ∀r>0, Sr(x)∩(A∪B) is not empty (so the statement similar to A is true) but that the statement similar to B is not true. That is, Sr(x)∩A is empty for a collection of r's which →0, and Sr(x)∩B is also empty for a collection of r's which →0. You can't "fix" this without some ideas about how the Sr(x) sets behave together. The necessary ingredients are present when we use Nr(x), but are not present more generally.
b) (2 points) A valid example.
c) (2 points) A valid example.

Problem 4 (15 points)
a) (13 points) 3 points for the easy part (that the diameter of the set is ≤ the diameter of the closure). I reserve 8 of the remaining 10 points for really attacking the sup correctly.
The statement that there exist x* and y* in the closure of A which actually have distance between them equal to the diameter of the closure is false even if all the diameters are finite. Let's get an example. Take X (the metric space, our "world" for this consideration) to be (0,1), the open unit interval in R. Let A be the subset of rational numbers in X. Then the diameter of A is 1. The closure of A is all elements of X by the density of the rationals in the reals. But notice that although the diameter of A and the diameter of the closure of A are both 1 (they are equal, of course!) there are no points x* and y* in (0,1) with d(x*,y*)=|x*-y*|=1. I deducted 2 points for students who "used" the "fact" that extremizing points exist. Most of those students presented a proof of the desired result which could be fixed up and validated without much more work!
b) (2 points) A valid example.

Problem 5 (15 points)
a) (13 points) This can be a very frustrating problem. I gave two proofs on the answer sheet. The doubling part in the cover proof (the first proof) reflects a certain amount of experience. One student wrote Argh! upon realizing that "just" taking the minimum radius of any finite subcover by balls won't quite work. I will assign 8 points to that work. The frustration involved in realizing that the original finite subcover doesn't work is common, and therefore mathematicians have assigned that frustration a name: the Lebesgue number. See here or here.
b) (2 points) A valid example.

Problem 6 (14 points)
a) (7 points) What's needed is an open cover without a finite subcover.
b) (7 points) 4 points for showing that an "infinite tail" of the sequence is inside one ball (2 of those points for citing the Cauchy property!), and 3 more for completing the proof.

Problem 7 (15 points)
A vicious, rude, horrible problem with nothing but inequalities. Why did so many people assume that the sequence is real? It could be a sequence of complex numbers. I will take off 2 points for a proof which relies on order. Let's see: 1 point for a "correct" (?) answer and 4 points for a valid use of the triangle inequality. Another 5 points given for some algebraic decomposition leading to a solution.

Maintained by and last modified 10/21/2008.