Here is how it is worded in Apostol:

Theorem. If a bounded set $S$ in $\mathbb{R}^n$ contains infinitely many points, then there is at least one point in $\mathbb{R}^n$ which is an accumulation point of $S$.

Proof. (for $\mathbb{R}^1$) Since $S$ is bounded, it lies in some interval $[-a, a]$. At least one of the subintervals $[-a, 0]$ or $[0, a]$ contains an infinite subset of $S$. Call one such subinterval $[a_1, b_1]$. Bisect $[a_1, b_1]$ and obtain a subinterval $[a_2, b_2]$ containing an infinite subset of $S$, and continue this process. In this way a countable collection of intervals is obtained, the $n$th interval $[a_n, b_n]$ being of length $b_n -a_n = a/2^{n-1}$. Clearly the sup of the left endpoints $a_n$ and the inf of the right endpoints $b_n$ must be equal, say to $x$. The point $x$ will be an accumulation point of $S$ because, if $r$ is any positive number, the interval $[a_n, b_n]$ will be contained in $B(x;r)$ as soon as $n$ is large enough so that $b_n -a_n < r/2$. The interval $B(x;r)$ contains a point of $S$ distinct from $x$ and hence $x$ is an accumulation point of $S$.

My questions:

1) Why bother with the first 3 sentences? What's wrong with just saying "Since $S$ is bounded, there is an interval $[a_1, b_1]$ containing an infinite subset of $S$". Actually, now I'm not entirely sure why it has to be bounded. Couldn't you always find some infinite subset to do this with?

2) Why does $b_n -a_n < r/2$? I would think it just has to be less than $r$. For example, these intervals could just "converge" to one end point ($x = a_1$ or $x = b_1$), in which case the length just has to be less than the radius of the ball (neighborhood).

3) I'm not entirely sure why we kept bisecting the intervals. I'm confused, but I don't really have a good concrete question.

  1. We need to start somewhere. So we start by noting $S\subset [-a,a]$ for some $a$. After we do this, we keep bisecting the interval ad infinitum. $[a_1,b_1]$ is just the result of the first bisection. You'll see why the interval needs to be bounded soon.

  2. $r$ here is just any positive number. Every time you bisect an interval, its length halves. $[a_n,b_n]$ stands for the interval you get after bisecting $n$ times, so its length is $\frac{2a}{2^n}$. Now, the point is that by increasing $n$, you can get this number to be as small as you want, since it converges to $0$. In particular, for large enough $n$, this length is less than $r/2$. The author then uses this to show that $[a_n,b_n]$ is contained in $B(x;r)$. (this also explains why the interval $[-a,a]$ we start out with should be bounded. Because we want its size to converge to $0$ when we bisect it; this wouldn't happen if we started with an infinite interval).

  3. The idea is: if we keep cutting the interval in half, each time choosing where we cut it so that it contains infinitely many points of $S$, then, as mentioned in 2., the interval's size will get smaller and smaller, so it will "zero in" on a single point. We call this point $x$, and this point is the accumulation point whose existence we set out to prove. It's intuitively clear why it's an accumulation point of $S$: because the intervals $[a_n,b_n]$ are smaller and smaller sets containing $x$, but also containing infinitely many points of $S$.

  • $\begingroup$ although I didn't post the question I thank you for the great answer! $\endgroup$ – DanZimm May 16 '13 at 3:12

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