From: Stephen Dalton 
Date: October 26, 2009 8:09:00 PM EDT
To: James Propp 
Subject: Comments on 92.305 Assignment #5

- Exercise 3.2.1: The proof referenced does hinge on the fact that the collection of epsilons be finite, but points were deducted for students who said finiteness was important because "we cannot take the minimum of a infinite set."  Check out the set of positive integers; it's infinite, for sure, but has a minimum.  Or take the interval [0,1].  Infinite?  Uncountably so!  And yet it has a clear minimum of zero.  The key to this problem was not that "if the set of epsilons were infinite we would never be able to find the minimum," but rather that "if the set of epsilons were infinite we MIGHT not be able to find the minimum."  Remember, if a general statement is to be discredited, all that is needed is one counterexample.  This "MIGHT not be able to" is enough to break a proof!

- 3.2.3: Infinity is not really a "limit point" of any subset of R.  Remember, a limit point of a set $A$ is a point $x$ that satisfies the condition that any open set containing $x$ will always intersect $A$ in some point other than $x$.  The problem here is subtle, but once stated it becomes clear, I think: infinity cannot be a limit point of R precisely because it is not really a real number, and therefore there is no such thing as "an open set containing $x$" in this case!  The symbol $\infty$ is written basically to denote that something never stops growing, not that it attains an "infinite" value.  This is suggested by the fact that unbounded intervals are always written with rounded brackets next to the infinity symbol; one writes $(0, \infty)$ and not $(0, \infty]$, indicating that infinity itself is not "included."

- A related point to what was just mentioned: a "limit point" is not the same as a "limit."  A "limit" is always something approached by a sequence, or a function, or a series, or some other thing that takes "steps" toward a "limit."  A set, which has limit *points* (and not "limits"), does not have this same notion of "progress" or "motion" as occurs with a sequence, a function, etc.  Here's one way, though, to understand the similarity in terminology.  First, a quick definition: by an "eventually constant" sequence, let's mean a sequence whose terms are eventually all the same; so the sequence (1,6,3,4,1,4,4,4,4,4,4,..) would be eventually constant, as would (1,1,1,1,...).  By a "never constant" sequence, let's denote a sequence that is NOT eventually constant; such a sequence might be (1, 1/2, 1/4, 1/8,...).

Now: if we have a set $A$, and we claim that $x$ is a limit point of $A$, this is the same thing as saying that we can construct a never-constant sequence $(a_n)$, with each term chosen from $A$, such that $(a_n)$ converges to $x$.  That is to say, $x$ being a "limit point" of $A$ means that it can be the "limit" of a sequence constructed entirely out of the elements of $A$, so long as those elements are not all the same (or even eventually all the same).  The proof that these two definitions are equivalent is not terribly difficult.  Why do we need the sequence $(a_n)$ to be "never constant?"  Consider the set $[0,1] \cup {3}$; if we allow a sequence of elements from this set to be eventually constant, then we would end up "proving" that 3 is a limit point of the set, which it is not.

- 3.2.3.e: Some students had trouble reading this problem.  Remember, sets are always written in the following manner:

{ (a symbol, or combination of symbols, to denote the object in question) : (a condition on that symbol) }

The set in this question is no different, although the "..." part probably hearkened back to infinite series and so made it confusing.  Parse it out, and you'll see that this is another way of writing:

{ the $n$th partial sum of $\sum{a_m}$ : $n$ is a positive integer }.

Hopefully, this makes it a little clearer.  So the set actually looks like this: {1, 1+1/4, 1+1/4+1/9, 1+1/4+1/9+1/16, ... }

- Ex 3.2.7.  Some students, after stating that "only finite many of the terms of $x_n$ lie outside of $O$," then proceeded to say something along the lines of "and infinitely many of the terms DO lie in $O$."  Really, the first is all that is needed.  A sequence has infinitely many terms, right?  Well, if "only finitely many" of the terms fail to satisfy a given condition, then the rest of the terms must satisfy that condition, right?  And "the rest of the terms" is infinitely many!  In fact, the statement that "only finitely many" FAIL to satisfy a certain condition is much stronger than the statement that "infinitely many" DO satisfy a given condition.  As an example, observe these two sequences:
1. (a_n) defined by (1,1,1/2,1,1/3,1,1/4,1,1/5,1,...)
2. (b_n) defined by (1,1/2,1/3,1/4,1/5,...)

Obviously, (b_n) converges to zero.  This is because given any epsilon-neighborhood of zero, *only finitely many* of the terms of (b_n) live outside the neighborhood.  Now look at (a_n).  It obviously DOESN'T converge to zero, because every other term is 1.  However, we could confidently state that given any epsilon neighborhood around zero, "infinitely many" of (a_n)'s terms lie in that neighborhood.  Thus, "only finitely many fail to satisfy (condition)" is much stronger than "infinitely many DO satisfy (condition)."

- Exercise 3.2.11: The supremum of a set is not always a limit point.  It's an interesting exercise to see when this IS the case, though.  In general, a set that contains its supremum will NOT necessarily have the supremum as a limit point, whereas a set that does NOT contain its supremum WILL necessarily have the supremum as a limit point.  (To see why this is true, toy around with the technical definitions of supremum and limit point.  Here's a hint: if a set $A$ contains its supremum $s$, then any epsilon neighborhood around $s$ will contain some element $a \in A$ that is greater than *or equal to* $s$.  Does this satisfy the conditions of $s$ being a limit point?)

- Exercise A: Reponses to this question were generally good, although many people lost a small amount of credit for failing to make the proof specifically about the Cantor set.  The intersection-of-closed-intervals method is definitely the way to go here, but full credit was awarded only to those students who specifically chose the closed intervals involved in the intersection to be iterations in the Cantor construction.  Just choosing them as arbitrary closed intervals kind of misses the point of the exercise.

-Steve