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, N_{r}(x)∪(A∩B) is *not* empty.

**B** At least one of the following statements is true:

∀r>0, N_{r}(x)∪A
is *not* empty.

∀r>0, N_{r}(x)∪B
is *not* empty.

Some students claimed that **A** and **B** are equivalent. This
*is* true because of the nested nature of N_{r}(x) as a
"function" of r. That is, if r_{1}<r_{2}, then
N_{r1}(x) is a subset of
N_{r2}(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 S_{r}(x) for r>0 to
mean the following:

If r∈[1/(n+1),1/n) and n is even then S_{r}(x)={0}.

If r∈[1/(n+1),1/n) and n is odd then S_{r}(x)={1}.

Notice that if A={0} and B={1}, then ∀r>0,
S_{r}(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, S_{r}(x)∩A is empty for a
collection of r's which →0, and
S_{r}(x)∩B is also empty for a
collection of r's which →0. You *can't* "fix" this without
some ideas about how the S_{r}(x) sets behave together. The
necessary ingredients are present when we use N_{r}(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
greenfie@math.rutgers.edu and last modified 10/21/2008.
**