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

## 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 :)

• 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). – Nicholas Roberts Mar 4 '20 at 16:26
• Cheers forgot about that. I'll add that now :) – George Cooper Mar 4 '20 at 16:29
• 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. – fleablood Mar 4 '20 at 17:41
• Cheers mate. Main concern was the fact it may be incorrect. Thanks for the response – George Cooper Mar 4 '20 at 17:43

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.
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 $$\lim_{k\to\infty}a_{n_k}=\alpha\neq\beta$$).