Countably Compact First-Countable Space is Sequentially Compact

From ProofWiki
Jump to: navigation, search

Theorem

A countably compact first-countable topological space is also 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 $T = \left({S, \tau}\right)$ 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 $\left \langle {s_n}\right \rangle$ be any sequence in $S$.

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

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

$\left\{{V_n: V_1 \supseteq V_2 \supseteq V_3 \supseteq \cdots}\right\}$

Then a subsequence $\left \langle {s_{n_i}}\right \rangle$, where $s_{n_i} \in V_i$, converges to $p$.

$\blacksquare$