Compact Subspace of Metric Space is Sequentially Compact in Itself
Theorem
Let $M = \struct {A, d}$ be a metric space.
Let $C \subseteq M$ be a subspace of $M$ such that $C$ is compact.
Then $C$ is sequentially compact in itself.
Proof
Let $C \subseteq M$ be compact.
Let $\sequence {x_n}$ be a sequence in $C$.
Let $S$ be the range of $\sequence {x_n}$.
Thus $S \subseteq C$ and $S$ may be either finite or infinite.
Finite Range
Let $S$ be finite.
Then at least one $x \in S$ must be repeated infinitely often in $\sequence {x_n}$.
Its occurrences form a subsequence converging to $x$.
$\Box$
Infinite Range
Let $S$ be infinite.
From Subsequence of Sequence in Metric Space with Limit, it is enough to show that $S$ has a limit point in $C$.
Aiming for a contradiction, suppose $S$ has no limit point in $C$.
Then for each $x \in C$, there exists $\epsilon \in \R_{>0}$ such that the open $\epsilon$-ball $\map {B_\epsilon} x$ contains no $y \in S: y \ne x$.
That is:
- $\map {B_\epsilon} x \cap S = \set x$
or
- $\map {B_\epsilon} x \cap S = \O$
Because $C$ is compact, the open cover $\mathcal B := \set {\map {B_\epsilon} x: x \in C}$ has a finite subcover.
But each $\map {B_\epsilon} x \in \mathcal B$ contains at most one point of $S$.
Therefore the union of any finite subset of $\mathcal B$ contains only finitely many points of $S$.
So no finite subset of $\mathcal B$ covers $S$, let alone $C$.
Thus $S$ must have a limit point in $C$.
$\blacksquare$
Sources
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $7.2$: Sequential compactness: Proposition $7.2.6$