I am having trouble on understanding how to prove compactness for a specific set in a Metric space Question: Let $(x_n) \rightarrow x_0$ in a metric space $(X, d)$ and $C = \{x_0, x_1, x_2, ...\}$.
(a)Prove that C is closed.
(b)Prove that C is compact in terms of open covers
proof:
(a)Since $C$ is the set that consists of only the sequence $(x_n)$, we know that any subsequence also converges to $x_0$. Using Theorem 3.2.5 from Abbott, we know that $x_0$ cannot be a limit point of C. Hence, $C$ has no limit points. Therefore, $C$ is closed.
I am having trouble proving that C is compact. I've tried $C \subseteq \bigcup U_i$ where $i \in I$. We want to show $\bigcup_{k=1}^n U_k$ where $U_k \subseteq U_i$
 A: Let $\{U_i\}_{i \in I}$ be an open cover of $C$. Then there's some $j$ s.t. $x_0 \in U_j$. Since $U_j$ is open, there's some $r >0$ with $B(x_0, r) \subset U_j.$
Now because $(x_n)$ converges to $x_0$, for all $\epsilon > 0$, there's some $N \in \mathbb N$ s.t. for all $n > N$ we have $|x_n - x_0| < \epsilon$. In particular, this is true for $\epsilon = r.$ Thus $x_n \in B(x_0, r)$ for all $n \ge N$ meaning $U_j$ contains $x_0$ and all the elements of $C$ past $N$ like $x_N, x_{N+1}, x_{N+2}, \ldots$
Up to $N$, there are only finite number of elements, that is, $x_1, x_2, x_3, \ldots, x_{N-1}$. Each one of these belongs to at least one $U_i$. To cover all these elements, we need at most $N - 1$ open sets ($U_i$s).
Thus we have shown the finite subcover $\{U_1, U_2, \ldots, U_{n - 1}, U_j\}$ covers $C$ and so $C$ is compact.
A: One useful way to say that $x_n\to x_0$ in a metric space is that for any open $U$ such that $x_0\in U,$ the set $\{n\in\Bbb N: x_n\not\in U\}$ is finite.
Let $G$ be an open cover of $C.$ Take $U\in G$ with $x_0\in U.$ Let $A=\{n\in\Bbb N: x_n\not\in U\}.$
Now the set $B=\{x_n:x_n\not\in U\}$ is finite because $A$ is finite and $\forall x\,(x\in B\iff \exists n\in A\,(x=x_n)).$ So take a finite set $H=\{g_x:x\in B\}\subset G$ such that $x\in g_x$ for each $x\in B.$
Now $\{U\}\cup H$ is a finite subset of $G$ and it covers $C$ because $\cup H\supset B=C\setminus U.$
