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I have the following problem:

Let $s$ be a real number and $(s_n)$ be a sequence of real numbers. Suppose that for any subsequence $(s_{n_{k}})$ of $(s_n)$, $(s_{n_{k}})$ has a subsequence $(s_{n_{k_{l}}}$) satisfying $$ \lim_{l \to \infty} s_{n_{k_{l}}}=s. \qquad (1)$$ Please show that $\lim\sup s_n = \lim \inf s_n = s$

I don't know if my proof is right. I have the following:

Let $S$ be the set of subsequential limits of $(s_n)$. Then, $S$ contains $s$ since it is the limit of some subsequence of $(s_n)$, namely $(s_{n_{k_{l}}})$, which is a subsequence of $(s_n)$. Now, I claim $S$ only contains $s$. Suppose, as a contradiction, that $S$ contains some other element $t$. That implies, that there exists a subsequence $(s_{n_{t}})$ with limit $t$. This implies that any subsequence of $(s_{n_{t}})$, namely $(s_{n_{t_{p}}})$ converges to $t$. But, that is a contradiction to the supposition that any subsequence of $(s_n)$ has a subsequence that converges to $s$. Hence, $S$ has only the element $s$. Therefore, $\lim\sup s_n=\sup S =s $ and $\lim\inf s_n=\inf S=s$, so $\lim\sup s_n = \lim \inf s_n = s$.

Would this be a complete valid proof?

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    $\begingroup$ Yes it is. Minor point: you don't have to say you're arguing by contradiction. What you show is: for any $t \in S$, in fact $t = s$. [For such a $t$, there's a subsequence $(s_{n_{k}}) \to t$, so any subsequence of that converges to $t$. But by hypothesis, some subsequence $(s_{n_{k_l}}) \to s$. As limits are unique, $s = t$.] Clearly, $s \in S$; so $S = \{s\}$. $\endgroup$
    – BrianO
    Oct 28, 2015 at 6:21

2 Answers 2

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Your idea is correct. But at the beginning of contradiction you should assume $t\neq s$. Then you should stated that that by assumption there exist $(s_{n_{t_{p}}}) \rightarrow s$ and therefore $s=t$ and this is the contradiction.

But maybe I am just too picky, good job!

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  • $\begingroup$ It's not necessary to assume $t \neq s$. Just, let $t$ be any element of $S$; then it turns out that $t = s$. Since $s \in S$, conclude $S = \{s\}$. $\endgroup$
    – BrianO
    Oct 28, 2015 at 6:13
  • $\begingroup$ Yes but then you do not have contradiction - as I said I had been picky, but my professors usually wanted to clearly stands, where the contradiction is. In your case you just proved, that for any element $t \in S$, $t=s$. That is completely correct, but then I would not use word contradiction in proof. $\endgroup$
    – iiivooo
    Oct 28, 2015 at 6:19
  • $\begingroup$ Well, sure: if you say you promise a contradiction, best to make clear what it is when you produce it :) I sketched my non-reductio ad absurdum idea in a comment to the original question. $\endgroup$
    – BrianO
    Oct 28, 2015 at 6:26
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Here is another way.

First show that $s_n \to s$:

Let $\epsilon>0$, and let $A = \{ n | |s_n-s| \ge \epsilon \}$. Then $A$ must be finite, otherwise we could find a subsequence satisfying $|s_{n_k}-s| \ge \epsilon$ for all $k$, which would contradict (1) immediately. Hence we see that $s_n \to s$.

Then show that if $s_n \to s$ that $\liminf_n s_n = \limsup_n s_n = s$:

Let $\epsilon>0$ and choose $N$ such that for $n \ge N$ we have $|s_n-s| < \epsilon$. In particular, we have $s-\epsilon < s_n < s+ \epsilon$ for $n \ge N$. Hence we have $s-\epsilon \le \inf_{n \ge N} s_n \le \sup_{n \ge N} s_n \le s+ \epsilon$ and hence $s-\epsilon \le \liminf_n s_n \le \limsup_n s_n \le s+ \epsilon$. Since $\epsilon>0$ is arbitrary, we have the desired result.

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