Compact Subspace of Metric Space is Sequentially Compact in Itself

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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