Local Compactness in a Topological Space

We have already described compactness of a topological space $X$, i.e., a set $A \subseteq X$ is said to be compact in $X$ if every open cover of $A$ has a finite subcover.

We will now describe the property of compactness locally.

Definition: A topological space $X$ is said to be Locally Compact at a point $x \in X$ if there exists a neighbourhood $U$ of $x$ that is compact, and $X$ is said to be locally compact (in general) if $X$ is locally compact on all of $X$.

For example, $\mathbb{R}^n$ with the usual topology is locally compact. To see this, let $\mathbf{x} \in \mathbb{R}^n$. Then for any $r > 0$, the closed ball centered at $\mathbf{x}$ with radius $r$ which we denote by $\bar{B}(\mathbf{x}, r)$ is a compact neighbourhood of $\mathbf{x}$.

To see this, note that $\bar{B}(\mathbf{x}, r)$ is closed and bounded, and a set in $\mathbb{R}^n$ is compact if and only if it is both closed and bounded!

Therefore $\bar{B}(\mathbf{x}, r)$ is indeed a compact neighbourhood of $\mathbf{x}$, so $\mathbb{R}^n$ is locally compact.

Of course, $\mathbb{R}^n$ itself is not compact, so the concept of compactness and local compactness are difference concepts.

For another example, let $X$ be any set with the discrete topology. Then for each $x \in X$ we have that the singleton set $\{ x \}$ is a neighbourhood of $x$ that is compact (since $\{ x \}$ is a finite set). So $X$ is locally compact. Of course, $X$ itself is compact if $X$ is a countable set, but $X$ is not compact if $X$ is an uncountable infinite set.

Nevertheless, the following proposition shows us that every compact space is also locally compact.

Proposition 1: Let $X$ be a topological space. If $X$ is compact then $X$ is locally compact.

• Proof: Suppose that $X$ is a compact space. Then for every $x \in X$, $U = X$ is a neighbourhood of $x$ that is compact. So $X$ is locally compact. $\blacksquare$