From: Stephen Dalton 
Date: November 5, 2009 1:21:39 AM EST
To: James Propp 
Subject: Comments on 92.305 assignment #6

- Record this on a tape and listen to it while you sleep: "The supremum and infimum of a set are not necessarily limit points of the set."  The definition of supremum and infimum of a set $A$ - that any neighborhood of them will necessarily contain some element $a \in A$ - only means that $\sup{A}$ is a limit point of $A$ if $sup{A}$ is not ITSELF this "element $a \in A$!"  Recall that a limit point $b$ of a set $B$ needs to have any neighborhood around $b$ intersect $B$ in some point *other than* $b$.  It was not acceptable to solve exercise 3.3.1 by appealing to the fact that $\sup{K}$ is a limit point of $K$, because in general, it is not.  What would be an acceptable way to solve the problem along these lines, though, is splitting into the two cases of $\sup{K} \in K$ and $\sup{K} \notin K$, and showing that in either case, $\sup{K}$ is in the closure of $K$.  (In the first case, it's in $K$ itself so of course it's in the closure.  In the second case, it can be shown that $\sup{K}$ is a limit point of $K$, but this is ONLY because the supremum is not actually IN the set!

- With set-theoretic proofs, be really careful that at no point in the proof are you making use of the truth of what you're trying to prove.  A few students lost some points for this.  When the problem consists of confusing symbols bouncing around, it can be hard to detect when you're committing this error of logic, but do be on the lookout for it.  One example of this was on exercise B, where some students at one point in their proof distributed an "interior" sign over an intersection - but the ability to do this is exactly what it is the goal of the problem to establish!

-Steve