3
$\begingroup$

To me, the problem is asking:

Assume $(a_n)$ is a bounded sequence. Prove:$[\text{Every convergent subsequence of }(a_n)\text{ converges to }a]\implies[\lim a_n=a]$

Step 0: We decide we'll prove the contrapositive, so the problem is now:

Assume $(a_n)$ is a bounded sequence. Prove: $[\lim a_n\ne a]\implies[\text{There exists a convergent subsequence of }(a_n)\text{ that does not converge to }a ]$

Step 1: $\lim a_n\ne a\implies\exists\ \epsilon_{0}>0:|a_n-a|\ge \epsilon_{0}\ \forall\ n\in\mathbb{N}$

Step 2: $(a_n)$ is bounded $\implies$ $(a_n)$ has a convergent subsequence. Call this convergent subsequence $(a_{n_{k}})$.

Step 3: The only things we know about $(a_{n_{k}})$ are that A) it converges to something and B) it's a subsequence of $(a_n)$. I think the only task left is to show $(a_{n_{k}})$ cannot converge specifically to $a$.

Step 4: To show $(a_{n_{k}})$ does not converge to $a$, we must show $\exists\ \epsilon_{1}>0:|a_{n_{k}}-a|\ge\epsilon_{1}\ \forall\ k\in\mathbb{N}$.

Step 5: $(a_{n_{k}})$ is a subsequence of $(a_n)$ $\implies$ All the terms of $(a_{n_{k}})$ come from $(a_{n})$. Since all terms of $(a_{n})$ are at least $\epsilon_{0}$ away from $a$, then so are all terms of $(a_{n_{k}})$.

Step 6: Let $\epsilon_{1}=\epsilon_{0}$. Then we have found an $\epsilon_{1}$ such that $|a_{n_{k}}-a|\ge \epsilon_{1}\ \forall\ k\in\mathbb{N}$.

Conclusion: $(a_{n_{k}})$ is a convergent subsequence of $(a_{n})$ (step 2) that does not converge to $a$ (steps 4-6). So we've found a convergent subsequence of $(a_{n})$ that doesn't converge to $a$.


I wrote this in these steps so it's easier to point out where I'm wrong. I don't understand why every solution I read (that doesn't involve limit superior) requires a sub-subsequence. It seems to me the reason the convergent sub-subsequence doesn't converge to $a$ is the same reason the convergent subsequence $(a_{n_{k}})$ doesn't converge to $a$, so why do we have to make a sub-subsequence?

I know this exact question has been asked on here ad nauseam. I read the posts I could find, but I still don't understand why a sub-subsequence is necessary. I'm self-learning, and I've been stuck on this for the past three days. I decided to finally just ask. I'm sorry.

$\endgroup$
3
  • 1
    $\begingroup$ Do note that $(a_n)$ is a subsequence of $(a_n)$. $\endgroup$
    – SvanN
    Aug 13 '17 at 13:44
  • $\begingroup$ Well, step 1 is already wrong. we could have $a_{2n}=a\,\forall n$ for example without that showing that $\lim a_n=a$. $\endgroup$
    – lulu
    Aug 13 '17 at 13:48
  • $\begingroup$ Counter-example: let $a_n =n $. The left hand side of the implication is vacuously true for all $a $. $\endgroup$
    – Myridium
    Aug 19 '17 at 20:58
4
$\begingroup$

Your proof is wrong, but it can be easily fixed. In step one, argue that there is a subsequence $(a_{n_k})_k$ of $(a_n)_n$ for which the conclusion in step $1$ holds. Then this subsequence will be bounded, so we can find a subsequence of this subsequence and make the same conclusions you did.

The reason this sort of argument requires a further subsequence is that just knowing that $(a_n)_n$ doesn't converge to $a$ means only that infinitely many terms of $a_n$ are away from $a$. However, in order to proceed with the proof, we need to have all terms away from $a$. Thus we need to take a subsequence first before proceeding with the proof - which will require a further subsequence.

In fact, something important regarding sequential convergence on topological spaces (or in particular, the real line) is at work here.

A sequence converges to $x$ iff every subsequence has a further subsequence that converges to $x$.

Proof: The forward direction is immediate. For the converse, suppose $x_n\not\to x$. Then there is a subsequence $(x_{n_k})_k$ of $(x_n)_n$ that stays $\varepsilon_0$ away from $x$. Hence no subsequence of $(x_{n_k})_k$ converges to $x$.

$\endgroup$
2
  • $\begingroup$ Oh I see, I've been misunderstanding the negation of $\lim a_n =1$ from the start. I didn't realise there could still be a few terms of $(a_n)$ close to $a$ (I thought $\lim a_{n}\ne a$ implied ALL of $(a_{n})$'s terms are $\epsilon_{0}$ away.) Thank you so much for help. $\endgroup$ Aug 13 '17 at 14:38
  • $\begingroup$ @anonanon444 A simple example to illustrate why this fails is the sequence $((-1)^n)_n$. This sequence doesn't converge to $1$ but it has infinitely many terms equal to $1$. $\endgroup$ Aug 13 '17 at 15:15
1
$\begingroup$

You can do without subsequences at all.

There is an $R\geq0$ such that $A:=\{a_n| n\geq1\}\subset B_R:=[{-R},R]$.

Let a neighborhood $U_\epsilon$ of $a$ be given. By assumption on the sequence $(a_n)_{n\geq1}$, for all $x\in B_R\setminus\{a\}$ there is a neighborhood $U_\delta(x)$ ($\delta$ may depend on $x$) such that $U_\delta(x)\cap A\subset\{x\}$. The family $\{U_\epsilon(a)\}\cup \bigl\{U_\delta(x) \bigm| x\in B_R\setminus\{a\}\bigr\}$ is an open covering of $B_R$, hence will contain a finite subcovering. There will be only finitely many $U_\delta(x_i)$ in this subcovering, hence there is an $n_0$ such that all $a_n$ with $n>n_0$ are lying in $U_\epsilon(a)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.