Union of Image of Convergent Sequence and Limit in Topological Space is Compact
Jump to navigation
Jump to search
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.
$\blacksquare$