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 greenfie@math.rutgers.edu and last modified 10/21/2008.