Assume {$a_n$} is bounded sequence with the property that every convergent subsequent of {$a_n$} converges to the same limit a R show that {$a_n$} must converge to a.

I don't understand that why {$a_{n_j}$} cannot converge to a...

So if the sequence does't converge to a, then its' subsequence cannot converges to a ?

Also, in this problem

Does "{$a_n$} does not converge to a" means that {$a_n$} is divergent ? or just {$a_n$} is convergent but not converges to a

I am confused.

  • $\begingroup$ Show.... what ? As much as some here make huge efforts to guess things, not all can do that. $\endgroup$ – DonAntonio Oct 20 '16 at 23:56
  • $\begingroup$ Sorry about that. I was putting the questions but it said I cannot write down over 150 :/ $\endgroup$ – Kwangi Yu Oct 21 '16 at 0:06

I'll not look at the provided solution, so maybe you'll get a different insight.

It's quite easy if you know about limes superior and limes inferior.

Since the given sequence is bounded, both limits (inferior and superior) are finite. There is a subsequence converging to the limes inferior and one converging to the limes superior. By assumption, these converge to the same number, therefore $$ \liminf_{n\to\infty}a_n=\limsup_{n\to\infty}a_n $$ and so the sequence converges.

Without the above concept, you can do as follows. First there is a converging subsequence, since the given sequence is bounded (Bolzano-Weierstraß). So we can assume, by contradiction, that the sequence doesn't converge to $a$ (the limit of all convergent subsequences).

This can mean it doesn't converge at all or that it converges to somewhere else than $a$, but it's not relevant. The important thing is that

there exists $\varepsilon>0$ such that, for all $N$, there is $n>N$ with $|a_n-a|\ge\varepsilon$.

Now choose $n_0>0$ such that $|a_{n_0}-a|\ge\varepsilon$. Next choose $n_1>n_0$ such that $|a_{n_1}-a|\ge\varepsilon$ and go on.

More precisely, if you have already chosen $n_k$ such that $n_0<n_1<\dots<n_k$ and $|a_{n_k}-a|\ge\varepsilon$, choose $n_{k+1}$ such that $|a_{n_{k+1}}-a|\ge\varepsilon$.

Thus we have a built a subsequence, but it's not necessarily convergent. However, since it is itself a bounded sequence, it has a convergent subsequence. This further subsequence cannot converge to $a$, because each of its terms $a_{n_{k_l}}$ satisfies $|a_{n_{k_l}}-a|\ge\varepsilon$. Contradiction.


Note that the subsequence $(a_{n_j})$ in step 3 is not just any subsequence; it's the subsequence of entries satisfying $|a_{n_j}-a|\geq\epsilon$ (we know from step 2 that there are infinitely many such entries, so we won't run out of them when constructing this subsequence).

Since each entry of $(a_{n_j})$ is at least $\epsilon$ away from $a$, this subsequence cannot converge to $a$. In fact any subsequence of $(a_{n_j})$ (a sub-subsequence of the original sequence) still has this property, so it also cannot converge to $a$.

Here "not converging to $a$" means it might not converge, or might converge to something other than $a$.

Here is an example which might clarify. Imagine $a=0$, $\epsilon=1$ and the sequence $(a_n)$ is $$ (a_n)=\left(1,2,\frac12,1,2,\frac13,1,2,\frac14,\ldots\right) $$ The subsequence $(a_{n_j})$ might consist of all entries which are at least $1$ (we assumed there were infinitely many such entries): $$ (a_{n_j})=(1,2,1,2,1,2,\ldots) $$ Now we can choose a convergent subsequence of this: $$ (a_{n_{j_k}})=(2,2,2,\ldots). $$ Since all the entries are at least $1$, this can't converge to $a=0$. So we found a convergent subsequence which doesn't converge to $a$, giving a contradiction.

  • $\begingroup$ Is this right ? so if a does not converge to a, we cannot say that its subsequenes also does not converge to a. But if we make a's subsequence (ank) that does not converge to a, then its subsequence (ankj) must not converge to a ??????? $\endgroup$ – Kwangi Yu Oct 21 '16 at 0:29
  • $\begingroup$ By the way, what is the meaning of entries ? $\endgroup$ – Kwangi Yu Oct 21 '16 at 0:32
  • $\begingroup$ By entries of $(a_n)$ I just mean the numbers $a_n$ for various values of $n$. You are right that in general it's possible that $(a_{n_{j_k}})$ converges to $a$ even though $(a_{n_j})$ doesn't. But in this case we have $|a_{n_j}-a|\geq\epsilon$ for all $j$, so $|a_{n_{j_k}}-a|\geq\epsilon$ for all $k$, so $(a_{n_{j_k}})$ cannot converge to $a$. $\endgroup$ – stewbasic Oct 21 '16 at 0:39
  • $\begingroup$ Thank you I got it. Since I did not know what the entries means. I was confused. Thank you $\endgroup$ – Kwangi Yu Oct 21 '16 at 2:21

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