TITLE: Prove that every bounded sequence in $\Bbb{R}$ has a convergent subsequence
QUESTION [5 upvotes]: I am to prove that every bounded sequence in $\Bbb{R}$ has a convergent subsequence. I am stuck not knowing how and where to start.

REPLY [7 votes]: Dichotomy.

Since the sequence $(x_n)_n$ is bounded, every $x_n$ is in a bounded interval $I_0$. Split $I_0$ into two subintervals of half its length. At least one of them contains infinitely many terms of the sequence, call it $I_1$. Repeat. After $n$ steps, $I_n$ is an interval of length the length of $I_0$ divided by $2^n$, and $I_n\subset I_{n-1}$. Call $I_n=(a_n,b_n)$ and pick $x_{\varphi(n)}$ in $I_n$ with $\varphi(n)\gt\varphi(n-1)$ (crucially, this is always possible since $I_n$ contains infinitely many terms of the sequence). Then $(a_n)_n$ is nondecreasing, $(b_n)_n$ is nonincreasing, $b_n-a_n\to0$ and $a_n\leqslant x_{\varphi(n)}\leqslant b_n$, hence everything is set so that the subsequence $(x_{\varphi(n)})_n$ converges.