I have a question regarding one exercise in Stephen Abbotts' Understanding Analysis. The question is: Assume $(a_n)$ is a bounded sequence with the property that every convergent subsequence of $(a_n)$ converges to the same limit $a \in \mathbb{R}$. Show that $(a_n)$ must converge to $a$.

I was then presented this proof (from the solutions manual): Assume for contradiction that $(a_n)$ does not converge to $a$, then from the negation of convergence we have (I'm not quite sure in this part): $$\exists \epsilon>0, \forall N\in \mathbb{N}, \exists n\in \mathbb{N} : n\geq N \wedge |a_n-a|\geq \epsilon. $$

Using this we can construct a subsequence $(a_{n_j})$ that diverges from $a$ as follows: for arbitrary $N\in \mathbb{N}$, we can always find a, say, $n_1$ such that, $n_1 \geq N$ where $|a_{n_1}-a|\geq \epsilon$. Since $n_1$ itself is in $\mathbb{N}$, then we can again find a $n_2 \geq n_1$ so that $|a_{n_2}-a|\geq \epsilon$. And in general we can find a $a_{n_{j+1}}$ after choosing an appropriate $a_{n_j}$ so that $|a_{n_{j+1}}-a|\geq \epsilon$.

From the construction of such a subsequence, isn't the contradiction that $(a_{n_j})$ does not converge to $a$ already proof that $(a_n)$ should converge to $a$? I am asking because the proof that I've read continues to say that the Bolzanno-Weierstrass theorem can be used to get another subsequence from the constructed $(a_{n_j})$ which diverges from $a$, and that the proof ends there. I think I understand it, but I fail to see why it is necessary. Can somebody enlighten me on this?

  • $\begingroup$ Your negation is slightly off. It should be $$\exists\varepsilon>0\;\forall N\in\mathbb N\;\exists n\in\mathbb N (n\geqslant N \wedge |a_n-a|\geqslant\varepsilon) $$ (where $\wedge$ denotes logical conjunction). $\endgroup$ – Math1000 Jun 15 '15 at 17:28
  • $\begingroup$ Really? I thought that the negation of $\exists a\in A, P(a)$ is $\forall a\in A, not(P(a))$. And that the negation of $P \implies Q$ is $P \implies not(Q)$. $\endgroup$ – Kurome Jun 15 '15 at 17:38
  • $\begingroup$ $$P\implies Q \equiv \lnot P \vee Q$$ So $$\lnot (P\implies Q) \equiv \lnot (\lnot P\vee Q) \equiv P\wedge\lnot Q$$ $\endgroup$ – Math1000 Jun 15 '15 at 17:41
  • $\begingroup$ I am still of the opinion that mine is logically equivalent to yours, unless I am wrong of course: is it true that $\neg (P \implies Q) \equiv P \implies \neg Q$ ? $\endgroup$ – Kurome Jun 15 '15 at 17:50
  • $\begingroup$ Write out the truth tables ;) $\endgroup$ – Math1000 Jun 15 '15 at 18:22

Here is how I would approach the problem - since $(a_n)$ is a bounded sequence, there exist subsequences $(a_{n_k})$ and $(a_{n_l})$ such that \begin{align}\lim_{k\to\infty} a_{n_k} &= \limsup_{n\to\infty} a_n\\ \lim_{l\to\infty} a_{n_l} &= \liminf_{n\to\infty} a_n. \end{align} (It is a good exercise to prove the above.) By hypothesis, $(a_{n_k})$ and $(a_{n_l})$ have the same limit. Therefore $$\limsup_{n\to\infty} a_n = \liminf_{n\to\infty} a_n = \lim_{n\to\infty} a_n. $$

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    $\begingroup$ That's interesting, I know that $lim\ sup\ a_n = lim\ inf\ a_n =a$ if and only if $lim\ a_n=a$. I do not know yet how to prove the one above that. $\endgroup$ – Kurome Jun 15 '15 at 17:36
  • $\begingroup$ This is interesting but not really what I'm asking. $\endgroup$ – Kurome Jun 15 '15 at 18:33

If you want to do it from scratch, argue as follows:

If $\left \{ a_{n} \right \}$ does not converge to $a$, the there is an $\epsilon >0$ and a subsequence $\left \{ a_{n_{k}} \right \}$ such that for each $k>0$, $\vert a_{n_{k}} \vert >\epsilon$.

But $a_{n_{k}}\subseteq a_{n}$, so it is bounded because $a_{n}$ is. Therefore, by Bolzano-Weierstrass, it contains a convergent subsequence, say $a_{n_{k_{l}}}$ which cannot converge to $a$.

And now you have your contradiction.

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  • $\begingroup$ "If ${a_n}$ does not converge to a, the there is an $ϵ>0$ and a subsequence $a_{n_k}$ such that for each $k>0$, $|a_{n_k}|>ϵ$." Do you not mean $|a_{n_k}-a| \geq \epsilon$? $\endgroup$ – Kurome Jun 16 '15 at 0:14
  • $\begingroup$ I am also not sure why we have to explicitly state that $a_{n_{k_l}}$ does not converge to $a$, isn't it already enough that we know $a_{n_k}$ does not converge to $a$ to show that $a_n$ must converge to $a$?. $\endgroup$ – Kurome Jun 16 '15 at 0:21
  • $\begingroup$ if so, then take $\epsilon /2$ the inequality is now strict. so you may as weel assume it's strict to begin with. $\endgroup$ – Matematleta Jun 16 '15 at 0:24
  • $\begingroup$ but all you know about $a_{n_{k}}$ is that it is bounded. You are given that know every convergent subsequence has $a$ as its limit.but $a_{n_{k}}$ might not converge. Bolzano saves you though because you can extract a convergent sub-subsequence. $\endgroup$ – Matematleta Jun 16 '15 at 0:30
  • $\begingroup$ I'm sorry, it might be annoying to keep answering my questions, but here's another one: the construction of $a_{n_k}$ does not necessarily mean it diverges from $a$? $\endgroup$ – Kurome Jun 16 '15 at 2:59

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