Closedness of Compact Sets in a Metric Space

# Closedness of Compact Sets in a Metric Space

Recall from the Compact Sets in a Metric Space page that if $(M, d)$ is a metric space then a subset $S \subseteq M$ is said to be compact if for every open covering of $S$ in $M$ there exists a finite subcovering of $\mathcal F$ that covers $S$.

On the Boundedness of Compact Sets in a Metric Space page we saw that if $S \subseteq M$ is compact in $M$ then $S$ is also bounded. We will see that if $S$ is compact then $S$ is additionally closed.

Theorem 1: Let $(M, d)$ be a metric space and let $S \subseteq M$ be a compact subset. Then $S$ is closed. |

**Proof:**To show that $S$ is closed, we will show that $S^c = M \setminus S$ is open by showing that for any $x \in M \setminus S$ that there exists a ball centered at $x$ with some radius $r > 0$ such that $B(x, r) \subseteq M \setminus S$.

- Let $x \in M \setminus S$. For any $s \in S$, let $r_s = d(x, s) \: (*)$. Consider the ball centered at $s$ with radius $\frac{r_s}{2} > 0$, i.e., $B \left ( s, \frac{r_s}{2} \right )$. Then $x \not \in B \left (s, \frac{r_s}{2} \right )$.

- Let $\mathcal F$ be the collection of such balls for each $s \in S$:

\begin{align} \quad \mathcal F = \left \{ B \left ( s, \frac{r_s}{2} \right ) : s \in S, \: r_s = d(x, s) > 0 \right \} \end{align}

- Then $\mathcal F$ is clearly an open covering of $S$ since for every $s \in S$ we have that $s \in B \left ( s, \frac{r_s}{2} \right) \subseteq \bigcup_{s \in S} B \left ( s, \frac{r_s}{2} \right)$.

- Now since $S$ is a compact set, we have that there exists a finite open subcovering subset $\mathcal F^* \subseteq \mathcal F$ of $S$. Since $\mathcal F^*$ is finite, there exists $s_1, s_2, ..., s_p \in S$ where:

\begin{align} \quad \mathcal F^* = \left \{ B \left ( s_1, \frac{r_{s_1}}{2} \right ), B \left ( s_2, \frac{r_{s_2}}{2} \right ), ..., B \left ( s_p, \frac{r_{s_p}}{2} \right ) \right \} \end{align}

- And such that:

\begin{align} \quad S \subseteq \bigcup_{k=1}^{p} B \left ( s_p, \frac{r_{s_p}}{2} \right ) \end{align}

- Since $\left \{ \frac{r_{s_1}}{2}, \frac{r_{s_2}}{2}, ..., \frac{r_{s_n}}{2} \right \}$ is a finite set, there exists a minimum. Say:

\begin{align} \quad r_{\mathrm{min}} = \min \left \{ \frac{r_{s_1}}{2}, \frac{r_{s_2}}{2}, ..., \frac{r_{s_n}}{2} \right \} \end{align}

- For all $k \in \{1, 2, ..., p \}$, we claim that $B(x, r_{\mathrm{min}}) \cap B \left ( s_k, \frac{r_{s_k}}{2} \right ) = \emptyset$. Let's prove this. Suppose that $y \in B(x, r_{\mathrm{min}}) \cap B \left ( s_k, \frac{r_{s_k}}{2} \right )$. Then $y \in B(x, r_{\mathrm{min}})$ and $y \in B \left ( s_k, \frac{r_{s_k}}{2} \right )$. Since $y \in B(x, r_{\mathrm{min}})$ we have that $d(x, y) < r_{\mathrm{min}} \leq r_{s_k}$ for all $k \in \{1, 2, ..., p \}$. Furthermore, since $y \in B \left ( s_k, \frac{r_{s_k}}{2} \right )$ we have that $d(y, s_k) < \frac{r_{s_k}}{2}$. By the triangle inequality:

\begin{align} \quad d(x, s_k) \leq d(x, y) + d(y, s_k) < r_{\mathrm{min}} + \frac{r_{s_k}}{2} \end{align}

- Since $r_{\mathrm{min}} \leq \frac{r_{s_k}}{2}$ for all $k \in \{1, 2, ..., p \}$ we have that:

\begin{align} \quad \leq \frac{r_{s_k}}{2} + \frac{r_{s_k}}{2} = r_{s_k} \end{align}

- Therefore $d(x, s_k) < r_{s_k}$. But $d(x, s_k) = r_{s_k}$ by $(*)$, so it cannot possibly be t hat $d(x, s_k) < r_{s_k}$, i.e., we've arrived at a contradiction. Therefore the assumption that $y \in B(x, r_{\mathrm{min}}) \cap B \left ( s_k, \frac{r_{s_k}}{2} \right )$ was false. Hence $B(x, r_{\mathrm{min}}) \cap B \left ( s_k, \frac{r_{s_k}}{2} \right ) = \emptyset$.

- Therefore $B(x, r_{\mathrm{min}}) \not \in B \left ( s_k, \frac{r_{s_k}}{2} \right )$ for all $k \in \{1, 2, ..., p\}$, so $B(x, r_{\mathrm{min}})$ is not contained in any sets in the covering $F^*$ of $S$. So $B(x, r_{\mathrm{min}}) \subseteq M \setminus S$. Since $x \in M \setminus S$ was arbitrarily chosen, we see that for all $x \in M \setminus S$ that there exists a ball centered at $x$ that is fully contained in $M \setminus S$. Therefore $M \setminus S$ is open, and $(M \setminus S)^c = S$ is closed. $\blacksquare$