Suppose that the sequence $\displaystyle \{ x_n \} $ is in $\displaystyle \mathbb {R} $.

Prove that $\displaystyle \{ x_n \} $ converges if and only if it has a unique cluster point.

My proof:

Given $\displaystyle \epsilon > 0 $, find the cluster point $\displaystyle x \in \mathbb {R} $ such that there exist infinite values n with $\displaystyle | x_n - x | < \epsilon $

I know that I need to find a $\displaystyle N \in \mathbb {N} $ such that whenever $\displaystyle n \geq N $, we have $\displaystyle | x_n - x | < \epsilon $.

But which N do I pick that so I can be sure that all the n afterward would get the sequence going into x and not going somewhere else? (In which I assume since the sequence only has one cluster point, the other point cannot go anywhere but x)

For the converse, let $\displaystyle \epsilon > 0 $, find $\displaystyle N \in \mathbb {N} $ such that whenever $\displaystyle n \geq N $, we have $\displaystyle |x_n - x | < \epsilon $.

Then there exist infinitely values of n such that $\displaystyle | x_n - x | < \epsilon $, which means x is a cluster point.

Now, suppose to the contrary that there exist another point $\displaystyle y \in \mathbb {R}, y \neq x $ such that there exist infinitely values of n with $\displaystyle | x_n - y | < \epsilon $.

I want to find an N such that the sequence would be converging both to x and y, which is impossible, but how should I go about choosing it?

Thank you!