Prove Bolzano-Weierstrass using MCT I have been picking up real analysis again and trying to do one of the exercises in the book I'm using. I'm asked to provide an alternate proof of Bolzano-Weierstrass. If $\{a_j\}$ is a bounded sequence, let $b_j = \inf\{a_j, a_{j+1}, \ldots\}$. Clearly $\{b_j\}$ is an increasing sequence that's bounded above, so by MCT this sequence converges. How should I proceed? I thought about showing $\{b_j\}$ is a subsequence of $\{a_j\}$ but I feel like this may not necessarily be true.
 A: For $n \in \mathbf{N}$ denote the tail $$T_n:= \{a_i : i \geqslant n\}.$$ Here's a proof by two cases:


*

*(At least) one tail has no maximum: In that case you can easily find an increasing subsequence of $\{a_j\}$.

*All tails have a maximum: In this case an appropriately chosen subsequence of $\{\max T_n : n \in \mathbf{N} \}$ does the trick.
A: The following has been proved.
Theorem 1: Every bounded increasing sequence is a convergent sequence.
Using theorem 1 we proceed to
Theorem 2: Every bounded sequence has a convergent subsequence.
Proof
Let $(a_j)_{j \ge 1}$ be a bounded sequence and let  $b_j = \text{inf}\{a_j, a_{j+1}, \ldots\}$ for $j \ge 1$. By theorem 1 we can write
$$\tag 1  \lim_{j\to +\infty} \,b_j = \beta$$
By the definition of $\text{inf}$ we can certainly find a $n \ge 1$ such that $|a_n - b_1| \lt 1$. In fact, for any $k \ge 1$ we can find $n \ge k$ such that $|a_n - b_k| \lt \frac{1}{k}$.
Using recursion we can build a strictly increasing function $\sigma: \Bbb N \to \Bbb N$ such that
$$\tag 2|a_{\sigma(k)} - b_k| \lt \frac{1}{k} \text{ and } \sigma(k) \ge k \quad \text{for all } k \ge 1$$
So the subsequence $(a_{\sigma(k)})_{k \ge 1}$ also converges to $\beta$ (see for example this link). $\quad \blacksquare$
