Sequentially Compact Metric Space is Compact/Proof 2
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Theorem
A sequentially compact metric space is compact.
Proof
Let $M = \struct {A, d}$ be a sequentially compact metric space.
Let $\UU$ be any open cover of $M$.
By Lebesgue's Number Lemma, there exists a Lebesgue number for $\UU$.
Let $\set {x_1, x_2, \ldots, x_n}$ be a finite $\epsilon$-net for $M$, where $\epsilon$ is this same Lebesgue number.
This exists by Sequentially Compact Metric Space is Totally Bounded.
Let $\map {B_\epsilon} {x_i}$ be the open $\epsilon$-ball of $x_i$.
By definition of Lebesgue number, $\map {B_\epsilon} {x_i}$ is contained in some $U_i \in \UU$.
Since:
- $\ds M \subseteq \bigcup_{i \mathop = 1}^n \map {B_\epsilon} {x_i} \subseteq \bigcup_{i \mathop = 1}^n U_i$
we have a finite subcover $\set {U_1, U_2, \ldots, U_n}$ of $\UU$ for $M$.
Hence the result.
$\blacksquare$
Sources
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $7.2$: Sequential compactness: Theorem $7.2.13$