Apostol Bolzano-Weierstrass Theorem 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


*

*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? 

*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 $B(x;r)$?

 A: 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 $b_n - a_n < r$.
A: 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.
A: I have been looking for the answer of this question for a while and the only thing I could come up with was that if $\sup\{a_n\} \neq \inf\{b_n\}$, then one of the two cases below must hold true:

*

*$\sup\{a_n\} > \inf\{b_n\}$

*$\sup\{a_n\} < \inf\{b_n\}$
It is obvious that the first case is not true; if it were true, the sub-interval $[a_n,b_n]$ would be empty which contradicts the premise of the proof.
For the second case, if the strict inequality holds true, then $\exists x$ s.t $\sup\{a_n\} < x < \inf\{b_n\}$. This indicates that we can bisect the sub-interval $[a_n,b_n]$ into either one of the two intervals $[a_n,x]$ and $[x,b_n]$. The process of bisecting sub-intervals, apparently, must continue until both endpoints of the interval converge to a point, say to $x$. Thus, the converging point $x$ must satisfy the condition $\sup\{a_n\} = x = \inf\{b_n\}$
I'm not sure wether the reasoning above is the exact reasoning Apostol intended to imply, but this reasoning helped me understand a simple and naive intuition of this theorem.
