# Show that the set $\{x_n:n\ge1\}\cup\{x\}$ is compact

Let $(X,d)$ a metric space and $(x_n)_n\subset X$ a convergent sequence with limit $x=\lim_{n\to\infty}x_n$. Show that the set $\{x_n:n\ge1\}\cup\{x\}$ is compact.

Well, i'm with difficulties in start this question. I know that some set is compact (sequentially) if for any sequence in the set, there is a subsequence convergent in a point in the set. But i'm not understanding what the question wants and how to prove it.

Could someone help me? Even to start it?

• Well, since you mention using sequences, any sequence in that space either contains $x$ infinitely often or finitely many times (possibly zero). Can you prove that there's a convergent subsequence in either case? Sep 15 '16 at 21:31
• When I read $\{x_n:n\ge1\}\cup\{x\}$, the "$\cup$" had conspicuously less space to its left and right than what is standard. It turned out that that was because the $\{\text{curly braces}\}$ are outside of MathJax rather than inside it. I changed that. See my edit to the question. $\qquad$ Sep 15 '16 at 21:38
• @TedShifrin Thankyou to the hint, friend, but i didn't understand the diference between x infinitely often and finitely many times... Can you discuss more about? Sep 15 '16 at 22:19
• @MichaelHardy Oh, my friend. Thank You! I am newbie and desperate (LoL) But i understood the suggestion. Thank You. Sep 15 '16 at 22:21
• @Lucas, if there are infinitely many entries of $x$ in your sequence, you have an obvious subsequence to choose. If there aren't, any subsequence of $\{x_n\}$ must converge (to a point of your set), so find one as a subsequence of your sequence. Sep 15 '16 at 22:25

Let $A=\{x\}\cup\{x_n:n\ge 1\}$; you’re supposed to prove that if the sequence $\langle x_n:n\ge 1\rangle$ converges to the point $x$, then the set $A$ is compact. I would use the definition of compactness: show that if $\mathscr{U}$ is an open cover of $A$, then $\mathscr{U}$ has a finite subset that still covers $A$.
HINT: Since $\mathscr{U}$ covers $A$, there is some $U_0\in\mathscr{U}$ such that $x\in U_0$. $U_0$ is open, so there is an $\epsilon>0$ such that $B_d(x,\epsilon)\subseteq U_0$. Now use the fact that $\langle x_n:n\ge 1\rangle$ converges to $x$ to show that all but finitely many points of $A$ are in $U_0$. It takes only finitely many more members of $\mathscr{U}$ to cover those finitely many points.
• I don't think "hint" is the right word for what you wrote. It's not a hint; it's a sketch of the argument. $\qquad$ Sep 15 '16 at 21:39
• @BrianM.Scott hello again (you already answered me in another question)! Thankyou for the support. I'm trying to unveil what you said and look what i've got: $\langle x_n:n\ge 1\rangle$ converges to $x$ means that exists an $\epsilon_2>0$ and an $N\in\Bbb{N}$ such that ${x_n}\subset{B_{\epsilon2}(x)}$ $\forall{n}$; $n>N$. Being $\epsilon_2\leq\epsilon$, implies that $B_{\epsilon2}(x)$, which contains infinite points of $\langle x_n:n\ge 1\rangle$, after $N$ points, is contained in ${B_{\epsilon}(x)}\subset{\mathcal{U_0}}\subset{\mathcal{U}}$. ... Sep 15 '16 at 23:22