Show that if $X$ is compact and $x$ is the only point of accumulation of the sequence ${x_n}$ then $x_n$ converges to $x$.

Question

Show that if $X$ is compact and $x$ is the only point of accumulation of the sequence $\{x_n\}$ then $x_n$ converges to $x$.

How could I prove it, I know that for the convergence to be fulfilled I have to prove it by the definition of convergence, but since I use the hypothesis of being the only limit point.

Pd: Disculpen la traducción, mi inglés no es muy fluido.

Answer 1

HINT: Show that if $U$ is any open nbhd of $x$, there are only finitely many $n\in\Bbb N$ such that $x_n\notin U$. To do this, suppose that there are infinitely many $n\in\Bbb N$ such that $x_n\notin U$. There are two possibilities:

• There is some $y\in X\setminus U$ such that $\{n\in\Bbb N:x_n=y\}$ is infinite. Why is this impossible?
• $\{x_n:x_n\notin U\}$ is infinite. Then $\{x_n:x_n\notin U\}$ is an infinite closed set with no accumulation point (why?); is this possible in a compact space?

Answer 2

Suppose $(x_n)$ does not converge to $x$, then there is some open set $O$ containing $x$, such that $X\setminus O$ is infinite, i.e. $M:=\{n: x_n \notin O\}$ is an infinite set. If $A:=\{x_n: n \in M\}$ is finite, then some $p \notin O$ occurs infinitely many times, and we have a new accumulation point of $(x_n)$, a contradiction. So $A$ is infinite and thus has an $\omega$-accumulation point in the compact subset $X\setminus O$ and again we have a new (not $x$) accumulation point of $(x_n)_n$. So we always get a contradiction and $x_n \to x$ after all.

Answer 3

If $x_n$ doesn't converge to $x$, then there's an open set $U$ such that $x \in U$ and infinitely many elements of $x_n$ fall outside of $U$. But $X \setminus U$ is compact, so that means $\{ x_n \} \cap (X \setminus U)$ has an accumulation point as well (If $E$ is an infinite subset of a compact set $K$, then $E$ has a limit point in $K$), contradicting the hypothesis that $x$ was the only accumulation point.