Proving Convergence with subsequences $\liminf$ and $\limsup$

Question

Problem:

$\{r_n\}$ is bounded in a way that every convergent subsequence $\{r_{n_k}\}$ converges to $r$. Prove $\{r_n\}$ converges to $r$.

Thoughts:

Since every convergent subsequence converges to $r$, then $\{r_n\}$ converges to $r$ since

$$\liminf r_n = \lim r_{n_k}$$for some sequence and the same is true for $\limsup r_n$.

Let \begin{align}a_n &:= \sup(r_k: k \ge n)\\ b_n &:= \inf(r_k: k \ge n)\end{align}

Note that $b_n\le r_n \le a_n$. From the reasoning above $\{a_n\}$ and $\{b_n\}$ have limits and they will be the same.

It then follows that through squeeze lemma that

$$\lim b_n = \liminf(r_n) = \limsup(r_n) = \lim a_n$$ Thus proving that it indeed converges.

Do I have the right idea here or is it completely off?

Answer 1

Assume that $\{r_n\}$ does not converge to $r$. Then there exists $\epsilon > 0$ such that the sequence never gets confined to the interval $[r-\epsilon, r + \epsilon]$. This means that there are infinitely many $n$ such that $$r_n \in K := [I,S]\backslash[r-\epsilon, r + \epsilon],$$ where $I = \inf_{n\in \mathbb N} r_n$ and $S = \sup_{n\in\mathbb N} r_n$. By assumption, $K$ is compact, and so $r_n$ has an accumulation point in $K$, which contradicts the fact that every subsequence converges to $r$.

Your proof is correct, but I don't see the point of the bottom part. Why not stop before "Let ..."?