Proof of Bolzano-Weierstrass Theorem for Real Sequences

Let $(x_n)_{n=1}^\infty$ be a bounded sequence of real numbers. Then there exists $M>0$ such that

Hence every term of the sequence lies in the closed interval $[-M,M]$.

Set $I_1=[-M,M]$. Having chosen a closed interval

that contains infinitely many terms of the sequence, divide $I_k$ into the two closed halves

At least one of these halves contains infinitely many terms of the sequence; choose such a half and call it $I_{k+1}$. Then

$I_{k+1}$ contains infinitely many terms of the sequence, and its length is

Inductively we obtain a nested sequence of closed intervals $I_k=[a_k,b_k]$ such that each $I_k$ contains infinitely many terms of the sequence and

In particular, $|I_k|\to 0$ as $k\to\infty$.

We now choose a subsequence $(x_{n_k})$. Since $I_1$ contains infinitely many terms, choose $n_1$ so that $x_{n_1}\in I_1$. Suppose $n_1<\cdots<n_k$ have been chosen with $x_{n_j}\in I_j$ for $1\le j\le k$. Because $I_{k+1}$ contains infinitely many terms of the original sequence, there exists an index $n_{k+1}>n_k$ such that

Thus $(x_{n_k})$ is a subsequence of $(x_n)$, and for every $j\ge k$ one has $x_{n_j}\in I_k$ because $I_j\subseteq I_k$.

We claim that $(x_{n_k})$ is a Cauchy sequence. Let $\varepsilon>0$. Choose $K$ so large that $|I_K|<\varepsilon$. If $m,n\ge K$, then both $x_{n_m}$ and $x_{n_n}$ lie in $I_K$, so

Therefore $(x_{n_k})$ is Cauchy. By Every Cauchy Sequence of Real Numbers Converges, there exists a real number $L$ such that $x_{n_k}\to L$ in the sense of Limit of a Sequence of Real Numbers.

Hence $(x_n)$ has a subsequence converging to a real number. This proves the theorem.