Compact Sets in a Metric Space
 Table of Contents

# Compact Sets in a Metric Space

Recall from the Coverings of a Set in a Metric Space page that if $(M, d)$ is a metric space and $S \subseteq M$ then a cover or covering of $S$ is a collection of subsets $\mathcal F$ in $M$ such that:

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

Furthermore, we said that an open cover (or open covering) is simply a cover that contains only open sets.

We also said that a subset $\mathcal S \subseteq \mathcal F$ is a subcover/subcovering (or open subcover/subcovering if $\mathcal F$ is an open covering) if $\mathcal S$ is also a cover of $S$, that is:

(2)
\begin{align} \quad S \subseteq \bigcup_{A \in \mathcal S} A \quad \mathrm{where} \: \mathcal S \subseteq \mathcal F \end{align}

We can now define the concept of a compact set using the definitions above.

 Definition: Let $(M, d)$ be a metric space. The subset $S \subseteq M$ is said to be Compact if every open covering $\mathcal F$ of $S$ has a finite subcovering of $S$.

In general, it may be more difficult to show that a subset of a metric space is compact than to show a subset of a metric space is not compact. So, let's look at an example of a subset of a metric space that is not compact.

Consider the metric space $(\mathbb{R}, d)$ where $d$ is the Euclidean metric and consider the set $S = [0, 1) \subseteq \mathbb{R}$. We claim that this set is not compact. To show that $S$ is not compact, we need to find an open covering $\mathcal F$ of $S$ that does not have a finite subcovering. Consider the following open covering:

(3)
\begin{align} \quad \mathcal F = \left \{ \left( (0, 1 - \frac{1}{n} \right ) : n \in \mathbb{Z}, n \geq 1 \right \} = \left \{ \left ( 0, \frac{1}{2}, \right ), \left ( 0, \frac{2}{3} \right ) , \left ( 0, \frac{3}{4} \right ) , ... \right \} \end{align}

Clearly $\mathcal F$ is an infinite subcovering of $(0, 1)$ and furthermore:

(4)
\begin{align} \quad (0, 1) \subseteq \bigcup_{k=2}^{\infty} \left ( 0, 1 - \frac{1}{k} \right ) \end{align}

Let $\mathcal F^*$ be a finite subset of $\mathcal F$ containing $p$ elements. Then:

(5)
\begin{align} \quad \mathcal F^* = \left \{ \left (0, 1 - \frac{1}{n_1}\right ), \left (0, 1 - \frac{1}{n_2}\right ), ..., \left (0, 1 - \frac{1}{n_p}\right ) \right \} \end{align}

Let $n^* = \max \{ n_1, n_2, ..., n_p \}$. Then due to the nesting of the open covering $\mathcal F$, we see that:

(6)
\begin{align} \quad \bigcup_{k=1}^{p} \left ( 0, 1 - \frac{1}{n_p} \right ) = \left ( 0, 1 - \frac{1}{n^*} \right ) \end{align}

But for $(0, 1) \subseteq \left ( 0, 1 - \frac{1}{n^*} \right )$ we need $1 \leq 1 - \frac{1}{n^*}$. But $n^* \in \mathbb{N}$, so $n^* > 0$ and $\frac{1}{n^*} > 0$, so $1 - \frac{1}{n^*} < 1$. Therefore any finite subset $\mathcal F^*$ of $\mathcal F$ cannot cover $S = (0, 1)$. Hence, $(0, 1)$ is not compact.

Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License