Union of Image of Convergent Sequence and Limit in Topological Space is Compact

Theorem
Let $\struct {X, \tau}$ be a topological space.

Let $\sequence {x_n}_{n \in \N}$ be a convergent sequence such that:


 * $x_n \to x$

Then:


 * $\ds \set {x_n : n \in \N} \cup \set x$ is compact.

Proof
Let $\family {U_\alpha}_{\alpha \in A}$ be an open cover for $\ds \set {x_n : n \in \N} \cup \set x$.

Take $\beta \in A$ be such that $x \in U_\beta$.

Since $x_n \to x$, there exists $N \in \N$ such that $x_n \in U_\beta$ for $n \ge N$.

For each $n < N$, we can select $U_{\alpha_1}, \ldots, U_{\alpha_{N - 1} }$ such that $x_j \in U_{\alpha_j}$ for each $1 \le j \le N - 1$.

Then we have:


 * $\ds \set {x_n : n \in \N} \cup \set x \subseteq U_\beta \cup \bigcup_{j \mathop = 1}^{N - 1} U_{\alpha_j}$

So every open cover has a finite subcover.

So $\ds \set {x_n : n \in \N} \cup \set x$ is compact.