Compact Sets in a Metric Space are Closed and Bounded

Recall from the Closedness of Compact Sets in a Metric Space page and the Boundedness of Compact Sets in a Metric Space that if $(X, d)$ is a metric space then every compact set in $X$ is also closed and bounded. We summarize this in the theorem below.

Theorem 1: Let $(X, d)$ be a metric space. If $A \subseteq X$ is compact then $A$ is closed and bounded.

One may wonder if the converse of Theorem 1 is true. It is important to note that if we are considering the metric space of real or complex numbers (or $\mathbb{R}^n$ or $\mathbb{C}^n$) then the answer is yes. In $\mathbb{R}^n$ and $\mathbb{C}^n$ a set is compact if and only if it is closed and bounded.

In general the answer is no. There exists metric spaces which have sets that are closed and bounded but aren't compact.

Theorem 2: There exists a metric space $(X, d)$ that has a closed and bounded set that is not compact.

• Proof: Let $\displaystyle{X = \left \{ \frac{1}{n} : n \in \mathbb{N} \right \}}$ and let $d : X \to [0, \infty)$ be the discrete metric defined for all $x, y \in X$ by:

(1)

\begin{align} \quad d(x, y) = \left\{\begin{matrix} 0 & \mathrm{if} \: x = y\\ 1 & \mathrm{if} \: x \neq y \end{matrix}\right. \end{align}

• Clearly $X$ is closed as it is the wholeset.

• Let $x_0 \in X$. Then $X \subseteq B(x_0, 1)$ and so $X$ is bounded.

• We now prove that $X$ is not compact. For each $n \in \mathbb{N}$ let $\displaystyle{X_n = \left \{ \frac{1}{n} \right \}}$. Observe that each $X_n$ is open since for each $n \in \mathbb{N}$ then set $\displaystyle{B \left ( \frac{1}{n}, \frac{1}{2} \right ) \subseteq X_n}$. Moreover, $\displaystyle{X = \bigcup_{n=1}^{\infty} X_n}$ so $\{ X_n : n \in \mathbb{N} \}$ is an open cover of $X$. However, $\{ X_n : n \in \mathbb{N} \}$ cannot be made smaller and still cover $X$. So there exists an open cover of $X$ with no finite subcover - so $X$ is not compact. $\blacksquare$