> [!theoremb] Theorem > > Every bounded [[Sequence|sequence]] has a [[Limit|convergent]] subsequence. > > *Proof*. Let $(a_n)$ be a bounded sequence, so that $\exists M > 0$ where $|a_n| \le M$ for all $n \in \nat$. Bisect the interval $[-M, M]$ into $[-M, 0]$ and $[0, M]$. Then, one of these closed intervals must contain an infinite number of values in the sequence. Select this half and denote it as $I_1$. Then let $a_{n_1}$ be some point in the sequence satisfying $a_{n_1} \in I_1$. > > Having selected an $I_k$, bisect it to two closed intervals and select one containing an infinite values as $I_{k+1}$, and let $a_{n_{k + 1}}$ with $n_{k + 1} > n_k$ be a point in the sequence in $I_{k + 1}$. Then, we obtain > $ > I_1 \supseteq I_2 \supseteq I_3 \supseteq \cdots > $ > By the nested interval property of the [[Real Numbers|reals]], their intersection contains at least one point $x \in \real$. Since the length of $I_k$ is $M(1/2)^{k - 1}$, which converges to $0$. Choose $N$ such that $k \ge N$ implies that the length of $I_k$ is less than $\varepsilon$. As $x, a_{n_{k}} \in I_k$, $|a_{n_{k}} - x| \lt \varepsilon$.