I have the following problem:

I'm only inspecting sup $A$ = lim inf $s_n$ since I imagine the proof will be more or less the same for both, but I haven't made much progress on this so far.

For the set $S$ of all subsequential limits of $s_n$, I know lim inf $s_n$ = inf $S$, and that the Bolzano-Weierstrass Theorem tells me $s_n$ has a subsequence which converges to some(any?) point in $[0,1]$. I'm not sure if I'm starting to look in the right place and just don't see it yet, or if I'm still completely lost.

I'm not sure if this is correct, or if it matters at all, but for any $a$ in $A$ we have $s_n < a$ for finitely many $n$. Since $s_n$ has infinitely many terms does this imply there are infinitely many terms of $s_n$ greater than $a$?

Thanks for any and all help!


1 Answer 1


I don't think we need the Bolzano-Weierstrass theorem, but note that BW says that every bounded infinite sequence has a limit point, but it is the fact that $[0,1]$ is a closed set that implies that the limit point is in $[0,1]$.

Let $S = \{x\in \mathbb{R} : \exists\;\; s_{n_k}\; \text{such that}\; s_{n_k} \rightarrow x \}.$

Then $\liminf s_n = \inf S$ by definition, and you want to show that

$$ \inf S = \sup A. $$

To show that $\sup A \leq \inf S$, it is enough to show that $a \leq \inf S$ for every $a \in A$. If $a\in A$, then $a \leq s_n$ for all $n \geq N$ (for some $N$.) In words, this means that $a$ is eventually less than or equal to every term in the sequence. But then $a$ must also be less than or equal to every term of any subsequence $s_{n_k}$, which implies that $a \leq x$ if $s_{n_k} \rightarrow x$. Thus $a \leq x$ for any $x\in S$, which implies that $a \leq \inf S$, so $\sup A \leq \inf S$.

For the other direction, to show that $\inf S \leq \sup A$, suppose that $\sup A < \inf S$ and let $a_n$ be a sequence in $A$ such that $a_n \rightarrow \sup A$. Since each $a_n$ is eventually less than or equal to every term in $s_n$, we can choose a subsequence $s_{n_k}$ such that $s_{n_k} \rightarrow \sup A$. But then $\sup A \in S$, which contradicts that $\sup A < \inf S$.

  • $\begingroup$ Ok, thanks a bunch for the reply, sorry for the delayed response. I follow the first part, and all of the second part makes sense, until you say that we can choose a subsequence of Sn that approaches sup A. I am unable to see why that is? $\endgroup$ Nov 7, 2016 at 2:36
  • $\begingroup$ Because for each $a_n$ you can choose some $s_{k_n}$ such that $a_n \leq s_{k_n} \leq \sup A$. If that was not the case, then we would have $a_n \leq \sup A < s_m$ for all $s_m \geq a_n$, which would contradict the definition of $\sup A$. $\endgroup$
    – user333870
    Nov 7, 2016 at 4:56
  • $\begingroup$ Ah ok, that makes sense. Thank you very much :) $\endgroup$ Nov 7, 2016 at 5:11

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