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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 $nth$ 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 is it obvious that the $\sup$ of the left endpoint $a_n$ is equal to the $\inf$ of the right endpoint $b_n$? Also how does one consider the $\sup$ or $\inf$ of a single number?
  2. I don't understand why we need $b_n−a_n<r/2$. Instead, why do we need to halve $r$? If $b_n−a_n<r$ then wouldn't $[a_n,b_n]$ still be contained in $B(x;r)$?
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    $\begingroup$ Yes, you're correct that if $b_n-a_n<r$, then $[a_n,b_n]\subset B(x;r)$. For $b_n-x\le b_n-a_n$ and, similarly, $x-a_n\le b_n-a_n$. $\endgroup$ – Ted Shifrin Aug 22 '16 at 5:36
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    $\begingroup$ Since $[a_{n+1},b_{n+1}]$ is a subinterval of $[a_n,b_n]$ by construction, the sequences $a_n$ and $b_n$ are (bounded and) decreasing and increasing, respectively, so $\lim_n a_n=\sup_n a_n$ and $\lim_n b_n=\inf b_n$. Since $|b_n-a_n|=a/2^{n-1}$, we have $\lim_n b_n-a_n=0$ and so $\lim_n a_n=\lim_n b_n$. (We do not consider $\sup$ and $\inf$ of single points, but range over all $n$.) $\endgroup$ – Luiz Cordeiro Aug 22 '16 at 15:39
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He means $\sup\{a_n : n \in \mathbb N\}$, not the supremum of the single point $a_n$.

Note that $a_n$ is an increasing sequence, and $b_n$ is decreasing. Both sequences are bounded, because $[a_n,b_n] \subset [a_1,b_1]$ for every $n$. Therefore they both converge (specifically, $a_n$ converges to its supremum, and $b_n$ converges to its infimum). As the distance $b_n - a_n$ is shrinking to zero, they must converge to the same limit.

I assume he halves $r$ because his ball $B(x;r)$ is open, so not quite large enough to contain a closed interval of length $r$ if one of the endpoints of the interval is $x$.


Edit: actually, he used a strict inequality $b_n - a_n < r/2$, so indeed he could have used $b_n - a_n < r$.

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  • $\begingroup$ Thank you very much, You guys are fantastic! $\endgroup$ – Jonathan Duran Aug 23 '16 at 0:05
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Apostol wrote "the endpoints" instead of "the endpoint". So your first question is solved by observing that he was saying $$ \sup_{n \geq 1\ (\text{say})}a_{n} = \sup \{ a_{n} \mid n \geq 1 \} = \inf \{ b_{n} \mid n \geq 1 \} = \inf_{n \geq 1}b_{n}. $$

For the second question, that part of the proof does not imply anything that is a must. That choice of an upper bound for $b_{n}-a_{n}$ where $n >> 1$ is to show the reader that we do can do so-and-so.

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  • $\begingroup$ So the proof would have been fine if we chose $r$ instead of $r/2$? Thank you for your response! $\endgroup$ – Jonathan Duran Aug 23 '16 at 0:04
  • $\begingroup$ Yes, it would also be fine. $\endgroup$ – Megadeth Aug 23 '16 at 1:40

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