# Countably Compact First-Countable Space is Sequentially Compact

## Proof 1

Follows directly from:

Infinite Sequence in Countably Compact Space has Accumulation Point
Accumulation Point of Infinite Sequence in First-Countable Space is Subsequential Limit

$\blacksquare$

## Proof 2

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 $p$.

$\blacksquare$