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


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?


1 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 ..."?

  • $\begingroup$ @Roberto_Rasapopoulos The reason I did this in such a way was because I was told that a way to solve this was to start with "A sequence has a subseqence such that lim x$_{n_k}$ = lim sup x$_n$". And after I had shown that I was supposed to show that "Sequence converges iff lim sup x$_n$ = lim inf x$_n$". Is the first part that I had really all that I need? $\endgroup$
    – Sky
    Mar 16, 2017 at 13:56
  • $\begingroup$ Well, once you conclude $\{r_n\}$ converges to $r$, you are done. $\endgroup$ Mar 16, 2017 at 15:36

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