Sequentially Compact Metric Space is Compact/Proof 2

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A sequentially compact metric space is compact.


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 $U_i \in \UU$.


$\ds M \subseteq \bigcup_{i \mathop = 1}^n \map {B_\epsilon} {x_i} \subseteq \bigcup_{i \mathop = 1}^n U_i$

we have a finite subcover $\set {U_1, U_2, \ldots, U_n}$ of $\UU$ for $M$.

Hence the result.