If $\{x_n\}$ converges to x then $ \{x_n\}_{n∈\mathbb{N}}\cup \{x\}$ is a compact space

Question

Prove that if $(x_n)$ is a sequence in $(X,d)$ that converges to a point $x$, then $ \{x_n\}_{n∈\mathbb{N}}\cup \{x\}$ is a compact space.

What I have tried has been the following:

Using the following definition of compactness: A set X is said to be compact if, given an open cover of any X, there is a finite undercoating.

Given an open covering {$_$} of $\{_\}_{n∈\mathbb{N}}\cup \{\}$, one of them $_{_0}$ contains the limit point $$.

By definition of limit that open $_{_0}$ contains all the points of the sequence less a finite number; that finite number of points that remains outside will each be contained in an open of the coating and thus all these open together with $_{_0}$ form the finite undercoating.

But I don't know if my idea is correct or if that is enough, thank you.

Answer 1

This community wiki solution is intended to clear the question from the unanswered queue.

Yes, your proof is correct.