Countably Compact First-Countable Space is Sequentially Compact/Proof 2

Theorem

A countably compact first-countable topological space is also sequentially compact.

Proof

Let $T = \struct {S, \tau}$ be a countably compact first-countable topological space.

By definition of sequentially compact, it is sufficient to show that every infinite sequence in $S$ has a convergent subsequence.

Let $\sequence {s_n}$ be any sequence in $S$.

By Infinite Sequence in Countably Compact Space has Accumulation Point, $\sequence {s_n}$ has an accumulation point $p \in S$.

As $T$ is first-countable, $p$ has a countable local basis, say:

$\set {V_n: V_1 \supseteq V_2 \supseteq V_3 \supseteq \cdots}$

Then a subsequence $\sequence {s_{n_i} }$, where $s_{n_i} \in V_i$, converges to $p$.

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$\blacksquare$

Sources

• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $3$: Compactness: Countability Axioms and Separability