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$(X,d)$ is a metric space and $(x_{n})$ is a sequence in $X$ that converges to $x\in X$.

How to show $\;\{x\}\cup\big\{x_{n}: n=1,2,3,\ldots\big\}\;$ is compact ?

I'm clueless here. I only have the idea that a subset in a metric space is compact iff it is closed and bounded.

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    The question is unclear. What is $x_n : x = 1,2,3, ...$ supposed to mean? Do you mean $x_n: n = 1,2,3,...$? If yes, what is $x$? What are the conditions on the sequence $(x_n)$? Are you trying to prove something or to disprove something? – Marktmeister Jan 01 '23 at 19:34
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    In $\Bbb{R}$ with the usual metric, is ${0}\cup{1,2,3,....}$ bounded? . Also your "idea" that a closed bounded subset of a metric space is compact is false . For example in $l^{1}$ , the unit ball is not compact. ${(1,0,0,...) , (0,1,0,....),...}={e_{1},e_{2},...,}$ is a sequence such that it has no convergent subsequence. In any case, you can take any infinite set $X$ and equip it with the discrete metric . Then it is closed and bounded but cannot be compact. – Mr.Gandalf Sauron Jan 01 '23 at 19:36

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It is certainly not clear but I will assume that you wanted to additionally write $x=\lim_{n\to\infty} x_n$ and forgot it.

Firstly, as @Mr.GandalfSauro commented above, your idea is definitely false and I would say it is important to know it.

To show that $\{x\}\cup\{x_n:n\in \mathbb{N}\}$ is compact, we can use the definition of a compact set. A set $K\subset X$ is compact if and only if for every collection $\mathcal{C}$ of open sets in $(X,d)$ such that $$K\subset\bigcup_{A \in \mathcal{C}} A$$ there exists a finite subcollection $\mathcal{C}'\subset \mathcal{C}$ such that $$K \subset \bigcup_{A \in \mathcal{C}'}A.$$

Let us try to apply this. Let $\mathcal{C}$ be a collection of open sets of $(X,d)$ such that $$\{x\}\cup\{x_n:n\in\mathbb{N}\} \subset \bigcup_{A \in \mathcal{C}}A.$$ Notice that there exists $A_0 \in \mathcal{C}$ such that $x \in A_0$. Since $A_0$ is open and $x$ is the limit of $(x_n)_{n \in \mathbb{N}}$, there exists $N \in \mathbb{N}$ such that $$n \geq N \Rightarrow x_n \in A_0.$$ Furthermore, for each $j =1,\ldots,N-1$, there exists $A_j \in \mathcal{C}$ such that $x_j \in A_j$. Now, we just observe that $\{A_j : j=0,1,\ldots,N-1\}$ is a finite subcollection of $\mathcal{C}$ that covers $\{x\}\cup\{x_n:n\in\mathbb{N}\}$ and we are done.

lfsm
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