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.

  • 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

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 $x$.

  • $\begingroup$ 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

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 $U_\epsilon(a)$.


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