# Compact Sets in Countable Complement Space

## Theorem

Let $T = \left({S, \tau}\right)$ be a countable complement topology on an uncountable set $S$.

Then the compact sets of $T$ are exactly the finite subsets of $S$.

## Proof

From Finite Topological Space is Compact, any finite subset of $S$ is compact.

Aiming for a contradiction, suppose $H \subseteq S$ is an infinite compact set.

Take a (countably) infinite sequence $\left \langle {a_n}\right \rangle_{n \ge 0}$ of distinct elements of $H$.

Consider the open sets:

$V_m := S \setminus \left\{{a_{m+n}}\right\}_{n \ge 0}$

for $m \ge 0$, which satisfy $V_{m_1} \subset V_{m_2}$ if $m_1 < m_2$.

Then:

$\displaystyle H \subseteq \bigcup_{m \mathop \ge 0} V_m$

is an open cover of $H$.

Since $H$ is compact, it has a finite open subcover, say:

$\displaystyle H \subseteq \bigcup_{i \mathop = 0}^N V_{m_i}$

with $m_0 < m_1 < \cdots < m_N$.

But then:

$\displaystyle \bigcup_{i \mathop = 0}^N V_{m_i} = V_{m_N}$

and so:

$\displaystyle a_{m_N+1}, a_{m_N+2}, \ldots \notin V_{m_N} = \bigcup_{i \mathop = 0}^N V_{m_i}$

which implies $a_{m_N+1}, a_{m_N+2}, \ldots \notin H$, a contradiction.

Hence the result by Proof by Contradiction.

$\blacksquare$