Bolzano-Weierstrass Theorem
Theorem
If a given infinite set $E \subset \mathbb{R}$ is bounded, then there is an accumulation point $p \in \mathbb{R}$ of $E$.
Explanation
Or one could also say, ‘A bounded sequence has a converging subsequence.’ It’s important to note that $E$ does not need to be closed under the condition.
Proof
Part 1. $\displaystyle \bigcap_{n=1}^{\infty} I_{n} = \left\{ x \right\}$
Since $E$ is bounded by assumption, there exists a closed interval $I_{1} := [a,b]$ that satisfies $E \subset I_{1}$. Here, let’s take $I_{2}$ as an interval that has half the length of $I_{1}$ and contains infinitely many points of $E$. By repeating this process, we can create a nested interval $I_{n+1} \subset I_{n}$, and the length of $I_{n}$ becomes $\displaystyle d_{n} : = {{b-a} \over {2^{n}}}$.
Cantor’s Interval Theorem: For nested intervals $[a_{n}, b_{n}]$, if $\displaystyle \lim_{n \to \infty} (b_{n} - a_{n}) = 0$ then $\displaystyle \bigcap_{n=1}^{\infty} [a_{n}, b_{n}]$ is a singleton set.
Since $\displaystyle \lim_{n \to \infty} d_{n} = 0$, by Cantor’s Interval Theorem, $\displaystyle \bigcap_{n=1}^{\infty} I_{n} = \left\{ x \right\}$ [NOTE : $x \in E$ is not necessarily required]
Part 2. $x$ is an accumulation point of $E$
Considering an open set $O$ that contains $x$, due to the density of real numbers, there exists a $\varepsilon>0$ that makes $(x - \varepsilon , x + \varepsilon ) \subset O$. As previously shown, since $\displaystyle \lim_{n \to \infty} d_{n} = 0$, a natural number $n_{0}$ that satisfies $\displaystyle d_{n_{0}} = {{b-a} \over {2^{n_{0}}}} < \varepsilon$ will also exist so
$$ x \in I_{n_{0}} \subset (x - \varepsilon , x + \varepsilon ) \subset O $$
Accumulation point: For a point $x \in \mathbb{R}$ in real numbers and a subset $A \subset \mathbb{R}$, if for any open set $O$ containing $x$, $ O \cap ( A \setminus \left\{ x \right\} ) \ne \emptyset $ then $x$ is defined as an accumulation point.
No matter how $O$ is defined, as long as it is an open set containing $x$, $\left( O \setminus \left\{ x \right\} \right) \cap E \ne \emptyset$ is inevitable. Therefore, $x$ becomes an accumulation point of $E$.
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