Compactness of Finite Sets in a Topological Space

# Compactness of Finite Sets in a Topological Space

Recall from the Compactness of Sets in a Topological Space page that if $X$ is a topological space then a set $A \subseteq X$ is said to be compact in $X$ if every open cover of $X$ has a finite subcover.

We will now look at a nice theorem that says that any finite set in any topological space is compact.

Theorem 1: Let $X$ be any topological space. If $A \subseteq X$ is a finite set then $A$ is compact in $X$. |

**Proof:**Let $A \subseteq X$ be a finite set, say:

\begin{align} \quad A = \{ x_1, x_2, ..., x_n \} \end{align}

- Let $\mathcal F = \{ A_i : i \in I \}$ be an open cover of $A$. Then:

\begin{align} \quad A \subseteq \bigcup_{i \in I} A_i \end{align}

- Since $\mathcal F$ covers $A$, the set $A$ can be partitioned into at most $n$ groups of elements from $A$ where all elements in these groups are contained in an open set in the collection $\mathcal F$. So there exists an subcollection $I^* \subseteq I$ where:

\begin{align} \quad A \subseteq \bigcup_{i \in I^*} A_i \end{align}

- Moreover, $\mid I^* \mid \leq n$. So $I^*$ is a finite subcover of $A$. Therefore $A$ is compact in $X$. $\blacksquare$

Corollary 1: If $X$ is a finite topological space then every subset $A \subseteq X$ is compact in $X$. |

**Proof:**Every subset $A$ of a finite set $X$ is finite. So by Theorem 1, $A$ is compact in $X$. $\blacksquare$