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Suppose that $a_n$ is a bounded sequence, with the property that there exists $l$ such that if any $a_{n_j}$ is any convergent subsequence of $a_n$ then its limit is $l$. Show that if n goes to infinity $a_n$ goes to l.

My main problem is that I don't think I understand the question.i cited it almost litteraly from my book. What I can make out of this is them asking us that if a subsequence of $a_n$ heads to limit $l$ then $a_n$ heads to $l$? However this is not true (take (0,1,0,1...).

So what do they want...?

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  • $\begingroup$ Does $a_n$ belong to a metric space? If so, think about the triangle inequality in respect to $a_n$ and two subsequences.. $\endgroup$ – MathematicianByMistake Mar 13 '16 at 14:38
  • $\begingroup$ @G.Sassatelli so why can't I just use the identity subsequence? And be immediately done? $\endgroup$ – Bahbi Mar 13 '16 at 14:40
  • $\begingroup$ @Bahbi I wrote an answer explaining what the problem asks (not the solution, of course). $\endgroup$ – user228113 Mar 13 '16 at 14:45
  • $\begingroup$ hint: Suppose $a_n \nrightarrow l $. Then there exists $\epsilon >0$ such that for each $N\in \mathbb N$...and since $a_n$ is bounded... $\endgroup$ – Matematleta Mar 13 '16 at 14:55
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In general, pick a non-convergent sequence $b_n$. Nothing prevents it from having convergent subsequences, so consider two convergent subsequences $b_{n_k}\to l_1$, $\ b_{n_h}\to l_2$.

Since $b_n$ is not convergent, a priori nothing guarantees that $l_1=l_2$. Moreover, since $b_n$ is not convergent, it might have non-convergent subsequences as well (namely, $b_n$ itself).

The exercise asks you to show that, if all convergent subsequences of a bounded sequence $a_n$ share the same limit $l$, then $\lim\limits_{n\to\infty} a_n=l$.

Your "counter-example" $a_n=\begin{cases}0&\text{if }n\text{ is even}\\1&\text{if }n\text{ is odd}\end{cases}$ is not a counter-example at all, since you can find two convergent subsequences of $a_n$ with different limits: $$\lim_{n\to\infty}a_{2n}=0\\\lim_{n\to\infty}a_{2n+1}=1$$ The point is showing that such a thing fails to happen only when the whole sequence $a_n$ was convergent to begin with.

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    $\begingroup$ you do need boundedness: take $a_n=1/n$ if n is even, and $a_n=n$ if not. $\endgroup$ – Matematleta Mar 13 '16 at 14:57
  • $\begingroup$ Thanks, @Chilango: sometimes I have "$\lim_{n\to\infty}a_n=+\infty$" in the back of my mind, leading me to silly mistakes. $\endgroup$ – user228113 Mar 13 '16 at 14:59

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