Boundedness of Compact Sets in a Metric Space

# Boundedness 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 in $M$ if for every open covering of $S$ there exists a finite subcovering of $S$.

We will now look at a rather important theorem which will tell us that if $S$ is a compact subset of $M$ then we can further deduce that $S$ is also a bounded subset.

Theorem 1: If $(M, d)$ be a metric space and $S \subseteq M$ is a compact subset of $M$ then $S$ is bounded. |

**Proof:**For a fixed $x_0 \in S$ and for $r > 0$, consider the ball centered at $x_0$ with radius $r$, i.e., $B(x_0, r)$. Let $\mathcal F$ denote the collection of balls centered at $x_0$ with varying radii $r > 0$:

\begin{align} \quad \mathcal F = \{ B(x_0, r) : r > 0 \} \end{align}

- It should not be hard to see that $\mathcal F$ is an open covering of $S$, since for all $s \in S$ we have that $d(x_0, s) = r_s > 0$, so $s \in B(x_0, r_s) \in \mathcal F$.

- Now since $S$ is compact and since $\mathcal F$ is an open covering of $S$, there exists a finite open subcovering subset $\mathcal F^* \subset \mathcal F$ that covers $S$. Since $\mathcal F^*$ is finite, we have that:

\begin{align} \quad \mathcal F^* = \{ B(x_0, r_1), B(x_0, r_2), ..., B(x_0, r_p) \} \end{align}

- And by definition $\mathcal F^*$ covers $S$ so:

\begin{align} \quad S \subseteq \bigcup_{k=1}^{p} B(x_0, r_k) \end{align}

- Each of the open balls in the open subcovering $\mathcal F^*$ is centered at $x_0$ with $r_1, r_2, ..., r_p > 0$. Since the set $\{ r_1, r_2, ..., r_p \}$ is a finite set, there exists a maximum value. Let:

\begin{align} \quad r_{\mathrm{max}} = \max \{ r_1, r_2, ..., r_p \} \end{align}

- Then for all $k \in \{ 1, 2, ..., p \}$ we have that $B(x_0, r_k) \subseteq B(x_0, r_{\mathrm{max}})$ and therefore:

\begin{align} \quad S \subseteq \bigcup_{k=1}^{p} B(x_0, r_k) = B(x_0, r_{\mathrm{max}}) \end{align}

- Hence $S$ is bounded. $\blacksquare$