Compactness of Sets in a Topological Space

# Compactness of Sets in a Topological Space

Recall from the Covers of Sets in a Topological Space page that if $X$ is a topological space and $A \subseteq X$ then a collection of subsets $\mathcal F$ of $X$ is said to be a cover of $A$ if:

(1)
\begin{align} \quad A \subseteq \bigcup_{U \in \mathcal F} U \end{align}

Moreover, we notes that if $\mathcal F$ is a collection of open sets satisfying the inclusion above, then $\mathcal F$ is an open cover of $A$, and analogously, if $\mathcal F$ is a collection of closed sets satisfying the inclusion above, then $\mathcal F$ is a closed cover of $A$.

We said that if $\mathcal F^* \subseteq \mathcal F$ also covers $A$ then $\mathcal F^*$ is a subcover of $A$.

We will now define a new type of set in a topological space known as compact sets.

 Definition: Let $X$ be a topological space and let $A \subseteq X$. Then $A$ is said to be Compact in $X$ if every open covering of $A$ (from $X$) has a finite open subcover.

As the reader may already be aware of, every closed and bounded interval $[a, b] \subset \mathbb{R}$ (where $\mathbb{R}$ has the usual topology) is compact. In fact, a set $A \subset \mathbb{R}$ is compact if and only if $A$ is both closed and bounded.

More generally, if $\mathbb{R}^n$ has the usual topology then $A \subseteq \mathbb{R}^n$ is compact if and only if $A$ is both closed and bounded.

In general, it’s rather difficult to show that a set is compact and it is usually easier to show that a set is not compact. For example, consider the set of natural numbers $\mathbb{N} \subset \mathbb{R}$. We claim that $\mathbb{N}$ is not compact in $\mathbb{R}$.

To show this, consider the following open cover of $\mathbb{N}$:

(2)
\begin{align} \quad \mathcal F = \left \{ \left (n - \frac{1}{2}, n + \frac{1}{2} \right ) : n \in \mathbb{N} \right \} \end{align}

Then each element in $\mathcal F$ contains exactly one natural number and the collection $\mathcal F$ contains a countably infinite collection of elements. If $\mathcal F^* \subset \mathcal F$ then:

(3)
\begin{align} \quad \mathbb{N} \not \subseteq \bigcup_{n \in \mathbb{N}} \left ( n - \frac{1}{2}, n + \frac{1}{2} \right ) \end{align}

This is because if $\mathcal F^* \subset \mathcal F$ then there exists an $n \in \mathbb{N}$ that is not contained in any of the open sets in $\mathcal F^*$ so $\mathcal F^*$ cannot cover $\mathbb{N}$. So, we have found an open cover $\mathcal F$ of $\mathbb{N}$ that has no finite subcover. Hence $\mathbb{N}$ is not compact in $\mathbb{R}$.