# 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). Mar 4, 2020 at 16:26
• Cheers forgot about that. I'll add that now :) Mar 4, 2020 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. Mar 4, 2020 at 17:41
• Cheers mate. Main concern was the fact it may be incorrect. Thanks for the response Mar 4, 2020 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 $$\lim_{k\to\infty}a_{n_k}=\alpha\neq\beta$$).