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HW1.
problem 7 - 2pts
21 - 2 pts
2 - 3pts
A1 - 3 pts
6 - 3pts
13 - 2pts

A lot of students struggled with problem 13, in particular that the
elements of F(N) are themselves finite subsets of N.  It doesn't matter if
those subsets are countable (they certainly are because they are finite),
it matters whether F(N) is countable.

In their proofs many people wrote down a surjection N^k onto S_k, where S_k
is the set of subsets of N consisting of k elements.  They mapped (a_1,
a_2, ..., a_k) onto {a_1, a_2,..., a_k}.  But this is not good because for
example (a,a,...., a) goes onto the set {a} which is not in S_k if k>1.

HW2.
Scoring breakdown:
Problem # - Points
4 - 3
12 - 2
5 - 2
11 - 2
2 - 3
B1 - 3

I saw a lot of people throwing around statements about sup(S) being an
element of the set S.  This is not true in general (for example sup([a,b))
= b, but b is not in [a,b) ).

Many students had trouble with B1.  There were two parts to the problem:
FIRST you need to prove that each of the sets [a,a], (a,b), [a,b), (a,b],
[a,b], (-\infty, b], etc... satisfy the definition of "interval."  This is
the easy part and was worth one point, but most people stopped there.
SECOND you need to prove that every interval is one of these sets [a,a],
(a,b), [a,b), (a,b], [a,b], (-\infty, b], etc....  To do this, you need to
break it into cases such as (1) unbounded above and below -- then it's
(-\infty,+\infty); (2) bounded above but unbounded below --- then subcases
of whether (2a) it contains its sup, so then it's (-\infty, b] or (2b)
doesn't contain its sup, and then it's (-\infty,b].

HW3.
The average homework score for HW#3 was 9.55/15.
The highest score was 13/15.
Point breakdown is as follows:
All problems were worth three points.
Problem 8 - proof was worth 1 pt, examples were worth 1 pt each because.
Problem 6 - Showing A is not closed was worth 1 pt, showing AU{0} is closed
was worth 2 pts.
Problem B2 - Showing d is a metric was worth 1 pt, showing that the disk is
open was worth 2 pts.

Problem 6 comments:
Proving that A is open got you no points.  This is because A is not in fact
open, and because showing a set is open doesn't mean it's not closed.

A SET CAN BE NEITHER OPEN NOR CLOSED, and it can be both open and closed
(although in the real numbers, the only sets that are both open and closed
are the whole real line and the empty set).

I saw a lot of people say that the complement of A is (-infinity, 0] U
[1,infinity).  But you're forgetting about all the little intervals in
(0,1) between all the numbers 1/n !!!

Problem B2 comments:
Most people did not understand what the problem was asking you to do.
...

HW4.
MATHEMATICAL WRITING ADVICE:
The homeworks are your chance to practice writing clear and concise proofs
that make sense.  Unlike the exams, you have several days to do the
homeworks, so you should use it as a chance to practice.  Here's one way to
do that.  Finish your homework a day or two days before it's due.  Leave
yourself time before you turn it in *and after you've had some time away
from it* like a good night sleep, other homework, etc. to read it over and
ask yourself "what the hell was I saying?"  "does this make sense to me
now?" "would someone else know what I was saying?" And if the answer is no,
fix it!  Chances are you'll understand the material a lot better once
you're done and you'll also identify your weaknesses.

All problems were worth 3 points.  I took off points for unclear epsilon
proofs.

In problem B3 there were many people who failed to write out a rigorous
proof.  You should start your solution like this:  "Assume the sequence has
a limit, call it x.  According to the definition of limit, for every
epsilon > 0 there exists a K such that |x_n - x| < epsilon for all n > K"
From there it's pretty easy to get contradictions in various ways.

For example, if you take epsilon < 1/2 in the first one, you can get a
contradiction pretty quickly based on what the sequence is.  (BUT you had
to write it out clearly!)

In the second one you can take epsilon to be anything you want and since
the sequence is made up of the natural numbers, you'd quickly get that the
natural numbers are bounded, which is a contradiction.  (BUT you had to
write it out clearly).

HW5.
Point breakdown
12 - 2pts
16 - 2pts
6c - 2pts
14a - 2pts
18 - 3 pts
22 - 2pts
B4 - 2 pts

For problem 14(a) I saw many many people argue that n^{1/n^2} <
(n^2)^{1/n^2} and this bigger sequence is a SUBSEQUENCE of n^{1/n} whose
limit we know is equal to 1.  While this argument is correct, I think what
the textbook authors had in mind (and what would be simpler), is to look at
the exponent and say 1/n^2 < 1/n, so n^{1/n^2} < n^{1/n}.  This way you
don't have to think about subsequences.

If you couldn't figure out the solution for number 18, check out the proof
of Theorem 3.2.11 in the book.  Similar techniques can be used except with
1 < r < L, and you'll be showing the sequence grows without bound instead of
showing it goes to zero.

For problem B4 virtually nobody successfully proved the hint.  I think only
one person successfully showed that there exist A, B real numbers such that
A < C_n < B for all n.  It's important to note here that you need to find
constant un-changing numbers A and B, which have no dependence on the index
n.  (You can get these numbers by using the assumption that all the x_n's
are in the interval [a,b]).

HW6.
Point distribution- all questions were worth 3 pts.

On problem 4 there were three parts to the problem, each worth one point:
(a) showing the sequence is increasing (i.e. x_n is less than or equal
x_{n+1}) (b) showing the sequence is bounded.  At this point you know the
sequence converges but the most exciting part of the problem is (c) showing
the limit equals 2.  This is nontrivial, but easy once you look at the
example in the book.  Since lim(x_{n+1}) = lim(x_n) we get the equation x =
\sqrt(x+2) where x is the limit of the sequence.  Then you solve the
quadratic equation and find that x = 2.

On part (b) of 3.5 #2 a lot of people got stuck.  There are many ways to do
this since, the nicest solution that I saw was bounding it by a telescoping
sum that you can cook up out of the factorials.

HW7.
I noticed there were several groups of students with identical proofs.
Please always be careful while you are working with other people to not
copy anyone's work.  If someone shares an idea, make sure you understand
that idea yourself and write it up yourself.

Average score was 7.6/15, high score was 12/15.

Point breakdown-- all problems worth 3 points.

C3: Many people lost points for failing to use the definition of an
infinite series.  When you just argue that "all the terms cancel" in an
infinite telescoping series, you are not being rigorous, and it shows that
you might not know what an infinite series actually is.  In an exam, Prof.
Koch will be looking to see that you understand that an infinite series is
limit of its partial sums.  So when you want to prove something about an
infinite series, write down something about its partial sums, (i.e. the sum
from n=1 to n=N of ther terms x_n) and then take a limit.

A5: It seemed like many people were afraid to try this problem, but it's
not bad.  Proving the metric space axioms is routine, but was worth 2
points.  In terms of the convergence, fix epsilon>0 arbitrary, then take N
big enough that 1/N < \epsilon, then for all n>N we have d(x_n,\infinity) =
1/n < 1/N < \epsilon.

B5: Nobody solved this problem, though two people got 2 out of 3 pts on it.
The sequences a_n and b_n are monotone increasing and decreasing
respectively,  and a_n < x_n < b_0 and b_n > x_n > a_0, so they are bounded
above and below respectively, hence have limits a and b respectively.

The sequence (a_n - b_n) must converge to zero since you always keep
cutting your intervals in half, i.e.
|a_n - b_n| = 2^{-n}(3/2), where 3/2 is the length of the original interval
[a_0, b_0].  Thus a = b.

Since a_n is less than or equal to x_n is less than or equal to b_n, the
squeeze theorem gives us that x_n converges to a limit x with a = x = b.

f is continuous so lim f(a_n) = f(a)=lim f(b_n) = f(b)=lim f(x_n) = f(x).

Since f(a_n) is always less than or equal to zero, lim f(a_n) = f(a) is
always less than or equal to zero (can you see why?).
Since f(b_n) is always greater than or equal to zero, lim f(b_n) = f(b) is
always greater than or equal to zero (can you see why?)

Hence 0 is less than or equal to f(x) is less than or equal to 0.

HW8.
Point breakdown:
B6- 3pts
4.2#3- 2pts
5.1#5 - 3 pts
5.2#10 - 3pts
5.3#1 - 1 pt
5.3#4 - 3pts

I took off half a point on B6 for not showing that 3 - (2/x^3) maps the
interval [2,3] to the interval [2,3].  This is a very important condition
for the contraction mapping theorem to work, because without it you could
have, for example, the function f(x) = 0 which certainly contracts
but has no fixed points in [2,3].

Also most people didn't have any trouble with 5.3 #1, but nobody did it the
way I think the book intended you to do it, which was to just cite Theorem
5.3.9 (it literally follows immediately).  I guess ultimately it just comes
down to a continuous function f achieving a min value on a closed interval,
which nearly everyone wrote down.