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Question

Show that a sequence of real numbers converges if and only if it is bounded and has not more than one accumulation point


Proof:

Let $(a_n)_{n\in\mathbb N}$ be a sequence which converges to $\alpha$. Let $\beta \ne \alpha\space$. Where $\beta\space$ is another accumulation point. Hence there is a subsequence such that $\lim_{k\to\infty}a_{n_k}=\beta$.

For $\epsilon := \frac{\alpha+\beta}{2}\gt0$

$\exists N\in\mathbb N$ such that $|a_{n}-\alpha|\lt\epsilon\space$ $\forall n\gt N$ and $\exists K\in\mathbb N$ such that $|a_{n_k}-\beta|\lt\epsilon\space$ $\forall k\gt K$.

Now choose $k^*\in\mathbb N$ such that both $k^* \gt K$ and $n_{k^*}\gt N$:

$|\alpha -\beta|\le|\alpha-a_{n_k^*}|+|a_{n_k^*}-\beta|\lt2\epsilon=|\alpha -\beta|$ which is a contradiction so there is only one accumulation point.

Also, choosing $\epsilon$ to be some positive number;

Let $\epsilon =1$:

$\Rightarrow |a_n-\alpha|\lt 1\Rightarrow |a_n|-|\alpha|\le|a_n-\alpha|\lt 1 \Rightarrow |a_n|\lt |\alpha|+1$

So if $n\gt N$, then $|a_n|\lt 1+|\alpha|$

Now consider where $n\le N$. This is a finite set so there exists a maximum value, call it $∣a_p∣$, that is $\max{(∣a_1∣,∣a_2∣,...,∣a_p∣,...,∣a_N∣})=|a_p∣$.

Let $M=\max({|a_p∣, 1+|\alpha|})$

$\forall n$, $|a_n|\le M$.

Hence $a_n$ is bounded

$\therefore$ Since $a_n$ converges $\Rightarrow$ $a_n$ bounded and has not more than one accumulation point.


Comment

This a question I have been asked in Analysis I. The question asks to prove it in bother directions (if and only if). I am unsure on how to do this in a succinct manner. Any tips/alternative proofs are really appreciated :)

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  • $\begingroup$ For the boundedness part, you only showed that terms past the index $N$ are bounded. How about the terms with index $n < N$? (This is easy since it's a finite list). $\endgroup$ – Nicholas Roberts Mar 4 '20 at 16:26
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    $\begingroup$ Cheers forgot about that. I'll add that now :) $\endgroup$ – George Cooper Mar 4 '20 at 16:29
  • $\begingroup$ I think your proof is good and succinct enough. It's more important to be clear and to acknowledge you are not skipping steps or making unwarranted assumptions than to be succinct. $\endgroup$ – fleablood Mar 4 '20 at 17:41
  • $\begingroup$ Cheers mate. Main concern was the fact it may be incorrect. Thanks for the response $\endgroup$ – George Cooper Mar 4 '20 at 17:43
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You have correctly proven the easy direction: If $\lim_{n\to\infty} a_n=\alpha$ then the sequence $(a_n)_{n\geq0}$ is bounded and cannot have another accumulation point $\beta\ne\alpha$.

For the other direction we have to consider an arbitrary sequence $n\mapsto a_n\in{\mathbb R}$ which is bounded, i.e., $|a_n|\leq M$ for some $M$, and has at most one accumulation point. In this case it has exactly one accumulation point $\alpha\in [-M,M]$, since $[-M,M]$ is compact. If $\lim_{n\to\infty}a_n=\alpha$ is wrong then there is an $\epsilon_0>0$ such that there are arbitrarily large $n$ with $|a_n-\alpha|\geq\epsilon_0$. These bad $a_n$ would lie in the compact set $S:=[-M,M]\>\setminus\>]\alpha-\epsilon_0,\alpha+\epsilon_0[\>$ and therefore would have an accumulation point $\beta\ne\alpha$, contrary to assumption.

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  • $\begingroup$ Perhaps I am missing something, but it seems that your answer is incomplete. Is it possible for a bounded sequence to have no accumulation point? $\endgroup$ – user2661923 Mar 4 '20 at 20:10
  • $\begingroup$ @user2661923: You are right. I have edited my answer slightly. $\endgroup$ – Christian Blatter Mar 5 '20 at 8:51
  • $\begingroup$ perhaps the OP (and others) have the same questions that I do: why does a bounded sequence have to have at least one accumulation point? What does it mean for a set to be compact, and why do compact sets have at least one accumulation point? $\endgroup$ – user2661923 Mar 5 '20 at 9:18
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What you did is correct. Clearly, your proof can be shortened if you can use the fact that every subsequence of a convergente sequence also converges and it has the same limit as the original sequence (because then $\lim_{k\to\infty}a_{n_k}=\alpha\neq\beta$).

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