Q:

About proof of theorem on topology

Theorem. If a convergent sequence in a metric space has infinitely many distinct points, then its limit is a limit point of the set of points of the sequence.

In the previous image is the proof of a theorem, may question is about the information used to reach absurd, is told that since x is in the sequence, the sequence should have only finitely many distinct a points, but this necessarily true? If we have like all {$x_n$} being distinct a points, for example n being negative integers, and all {$x_n$} being x, for all n being natural numbers, this is a convergent serie with limit x, and with infinitely many diferentes points, even do the aforementioned proof keep being true?
How may I proof it?

A:

We have a sequence $(x_n)_n$ with limit $x$. This is a function, that assigns to each $n$ some point $x_n \in X$. The set of points is $A = \{x_n: n \in \mathbb{N}\}$, which is a different object, which can be finite. $x$ is not in the sequence per se, nor need it be in $A$. 
But suppose $x$ is not a limit point of $A$ (the term limit point suggests $x$ should be a limit point of $A$, and this is very often the case, except in a trivial case we will see now). Then there is some neighbourhood $S_\epsilon(x)$ of $x$ such that $S_\epsilon(x) \cap A = \{x\}$ or $\emptyset$. We can apply the definition of a convergent sequence for this $\epsilon$, and we get some $N$ such that for all $n > N$, we have that $x_n \in S_\epsilon(x)$. All such points are in $A$, so for all those $n > N$: $x_n = x$. So the sequence is a trivial convergent sequence: all its tail consists of values $x$. And so $A$ can be at most $\{x_!,x_2,\ldots,x_N, x\}$ and $A$ is finite. A finite set actually never has a limit point in a metric space. So $x$ is not a limit point of $A$ iff $A$ is finite.

