3.7Local Compactness

Updated 23 Aug 2026Chapter (PDF)

[3.7.1]Definition(Local Compactness)#

A space $X$ is said to be locally compact at $x$ if there exists some compact set $C \subseteq X$ that contains an open neighborhood of $x$ . If each point of $X$ is locally compact, $X$ is said to be locally compact.

[3.7.2]Example(Examples of Locally Compact Spaces)#

Any compact space $X$ is locally compact, since for each $x \in X$ , there is some open neighborhood $O$ of $x$ such that $x \in O \subseteq X$ , with $X$ being the compact superspace.
$\bR$ is locally compact. For each $x \in \bR$ , there is some $a, b \in \bR$ such that $x \in (a, b) \subseteq [a, b]$ , where $[a, b]$ is compact. The same is true of $\bR^n$ , where instead we can draw an open ball around a point $x$ that is contained inside a closed one.
Any ordered set $X$ is compact since every basis element is contained in a closed interval.
$\bR^\omega$ in its product topology is not locally compact, since none of its basis elements are contained in compact superspaces. Take an arbitrary basic, open set $U = (\prod_{k = 1}^n [a_k, b_k]) \times \bR \times \cdots$ . If $U$ were contained in a compact set, then its closure $\bar{U}$ would be compact as well, since closed subsets of compact spaces are compact. However, $\bar{U}$ is the finite product of closed intervals, which is also multiplied by infinitely many instances of $\bR$ , making it not compct.

Of course, there are weird and arcane topological spaces. One method of demystifying such spaces is by considering them as subspaces of nice topological spaces, in hopes that it will inherit some nice properties. For a “bad” space $X$ , we can consider an embedding $f: X \to Y$ into a “good” space $Y$ , such that $(X \cong f(X)) \subseteq Y$ . In reality, $f(X)$ is the true subspace of $Y$ , but since $X$ is homeomorphic to $f(X)$ , $X$ has the nice properties $f(X)$ would inherit as a subspace of $Y$ . Through this, we are able to study a bad space by observing it as a subspace of a good space.

So what constitutes a good space? The examples we will consider are metrizable spaces and compact Hausdorff spaces. Metrizable spaces are nice due to a rigid topology defined by their metric, as well as topological notions (such as limit points and closure) being able to be probed by sequences. Compact Hausdorff spaces are nice since infinite covers can be reduced to finite subcovers, so some properties are more tractable, as well as sequences having unique limit points and other goodies.

Subspaces of metrizable spaces are metrizable as well, so no new information can be gathered from this. While subspaces of compact Hausdorff spaces are Hausdorff, they do not need to be compact. For example, consider the compact Hausdorff space $[0, 1]$ and a subspace $(0, 1)$ , which is obviously not compact. In later sections we will discuss a more complete characterization of every space that can be embedded into a compact Hausdorff space, but for now we investigate a simpler question. We investigate what classes of spaces are such that adding one specific point can produce a compact Hausdorff space. We will see that it is indeed locally compact Hausdorff spaces: there is a special point one could adjoin to a locally compact Hausdorff space to produce a compact Hausdorff space. This process is called one-point compactification.

[3.7.3]Theorem(Characterization of Locally Compact Hausdorff Spaces)#

Let $X$ be a space. Then $X$ is locally compact Hausdorff if and only if there exists a space $Y$ such that:

$X$ is a subspace of $Y$ .
$Y \setminus X$ is a singleton.
$Y$ is compact Hausdorff.

The identified space $Y$ is unique up to homeomorphism. If there are two spaces $Y, Y'$ satisfying the aforementioned properties, then there is a homeomorphism from $Y$ to $Y'$ that fixes $X$ .

Proof.

We first verify uniqueness. Let $X$ be a LCH space, and suppose $Y, Y'$ are superspaces of $X$ that have the prescribed properties, in such a way that $Y = X \cup \{\infty\}$ and $Y' = X \cup \{\infty'\}$ . Define a map

h: Y \to Y' \quad h(x) = \begin{cases} \infty' & x = \infty, \\ x & \text{otherwise,} \end{cases}

a map that fixes $X$ and takes $\infty \mapsto \infty'$ . We show that $h$ is a homeomorphism.

Obviously, $h$ is a bijection. We show that $h$ maps open sets to open sets, and the reverse is true by symmetry. To get started, suppose $O \subseteq Y$ is open. If $\infty \notin O$ , then $O \subseteq X$ is open, and so $h(O) = O \subseteq X$ is also open. If $\infty \in O$ , then note that $C = Y \setminus O$ is a closed subset of $X$ , and so $C$ is compact (since closed subspaces of Hausdorff spaces are compact). Since $X \subseteq Y'$ , we have that $C$ is a compact, and thus closed, subspace of $Y'$ . Thus $h(O) = Y' \setminus C$ is open, making $h$ a homeomorphism.

Now we show that if $X$ is a LCH space, there is some point $\infty \notin X$ such that $Y = X \cup \{\infty\}$ is compact Hausdorff. We equip $Y$ with the topology comprised of all sets $U$ that are open subsets of $X$ and all sets $Y \setminus C$ such that $C$ is compact in $X$ . We need to show that this topology is valid, the topology on $X$ is the subspace topology from $Y$ , and that $Y$ is compact Hausdorff with this topology. The first two conditions are trivial to check.

We show $Y$ is compact: consider an open cover $\mathcal{A}$ of $Y$ . Note that there is some compact $C \subseteq X$ such that $Y \setminus C \in \mathcal{A}$ , since open sets in $X$ do not contain $\infty$ . Then note all other members of $\mathcal{A}$ , when intersected with $A$ , form an open cover for $C$ with open subsets of $X$ . Since $C$ is compact, a finite subcollection covers $C$ , and when adjoined to the open set $Y \setminus C$ , a finite subcollection of $\mathcal{A}$ covers $Y$ . Thus $Y$ is compact.

We show $Y$ is Hausdorff; consider $x \ne y \in Y$ . Separated neighborhoods exist if $x, y \in X$ since $X$ is Hausdorff, so WLOG suppose $y = \infty$ . Since $X$ is locally compact, there is a compact $C \subseteq X$ such that there is some open neighborhood $U \subseteq C$ of $x$ . Then $U, Y \setminus C$ are disjoint open neighborhoods of $x$ and $y$ respectively, and so $Y$ is Hausdorff.

Now we prove the converse: if such a space $Y$ exists that satisfies the following conditions, then $X$ is locally compact (Hausdorffness is immediate). Given $x \in X$ , we show $X$ is locally compact at $x$ . Choose disjoint open sets $U, V$ of $Y$ containing $x$ and $\infty$ . Then $C = Y \setminus V$ is closed in Hausdorff space $Y$ and thus compact, and $U$ is an open neighborhood of $x$ that lies in the compact space $C \subseteq X$ . Thus $X$ is locally compact.

If $X$ happens to be a compact space, then the $Y = X \cup \{\infty\}$ from the preceding theorem is not interesting. Note if $X$ is compact, we need that $Y \setminus X$ is open in our prescribed $Y$ , so the singleton $\{\infty\}$ is open, making $\infty$ an isolated point. If $X$ is not compact, then $Y \setminus X$ is not open, and so $\{\infty\}$ is not open. Thus since every open neighborhood of $\infty$ intersects $X$ nontrivially, making $\infty$ a limit point of $X$ . Thus $\bar{X} = Y$ for $X$ not compact, where the closure is relative to $Y$ . We call this process compactification.

[3.7.4]Definition(Compactification, One-Point Compactification)#

If $Y$ is a compact Hausdorff space and $X$ is a proper subspace of $Y$ such that $\bar{X} = Y$ , $Y$ is called the compactification of $X$ . If $Y \setminus X$ is a singleton, then it is called the one-point compactification.

[3.7.5]Example(Common Examples of Compactifications)#

Ultimately, we have shown that if $X$ is LCH and not compact, then it admits a one-point compactification. Note this means that $\bR$ admits the one-point compactification $\bar{\bR} = \bR \cup \{\infty\}$ , which can be readily seen to be homeomorphic to $\bR^1$ by stereographic projection. The same is true for $\bR^2$ and in general $\bR^n$ with $S^n$ . Note if we see $\bC \cong \bR^2$ , we get a one-point compactification $\bC \cup \{\infty\} \cong S^2$ , and we even call the one-point compactification of $\bC$ the Riemann sphere.

Note there is an alternative, more satisfying way to define local compactness. Typically, when a property is local about some $x \in X$ , it means that there exists a neighborhood about $x$ where the property is true. But our definition of local compactness is not framed in this way. Indeed, we will formulate an alternative definition that identifies behavior more local in nature. Note this characterization is only equivalent in the context of Hausdorff spaces, however.

[3.7.6]Theorem(Characterization of Local Compactness)#

Let $X$ be a Hausdorff space. Then $X$ is locally compact if and only if, for any $x \in X$ and an open neighborhood $U$ of $x$ , there is some open neighborhood $V \subseteq U$ of $x$ such that $\bar{V}$ is compact and $\bar{V} \subseteq U$ .

Proof.

$(\Longrightarrow)$ : Suppose $X$ is LCH. Let $x \in X$ and an open neighborhood $U \subseteq X$ of $x$ be arbitrary. Let $Y$ be the one-point compactification of $X$ , then note $C = Y \setminus U$ is closed, and thus compact, in the Hausdorff space $Y$ . Note that a compact subset of a Hausdorff space can be separated from a singleton outside the compact subset through open sets. Thus choose disjoint open sets $V, W$ such that $x \in V$ and $C \subseteq W$ . Thus $\bar{V}$ is closed and thus compact in the Hausdorff space $Y$ , and since it is disjoint from $C$ , we have $\bar{V} \subseteq U$ as desired.

$(\Longleftarrow)$ : Suppose this formulation is true. Given $x \in X$ and an open neighborhood $U \subseteq X$ , there exists a neighborhood $V$ of $x$ such that the compact space $\bar{V}$ is contained in $U$ . Since $\bar{V}$ is a compact subspace containing the open neighborhood $V$ of $x$ , it follows that $X$ is locally compact.

That is, a Hausdorff space $X$ is locally compact if and only if every neighborhood about any point $x \in X$ contains a compact subspace that contains an open neighborhood about $x$ .

[3.7.7]Corollary(Locally Compact Subspaces of LCH Spaces)#

Let $X$ be an LCH space. If $A \subseteq X$ is open or closed, then $A$ is locally compact.

Proof.

Suppose $A$ is closed, and let $x \in A$ . Note that $X$ is locally compact, so there is some compact set $C \subseteq X$ that contains an open neighborhood $U \subseteq X$ of $x$ . Then note $C \cap A$ is closed in $C$ and thus compact, and it contains the open neighborhood $U \cap A$ of $x$ in $A$ .

Now suppose $A$ is open, and let $x \in A$ . Since $X$ is locally compact and $A$ is a neighborhood of $x$ , there is some neighborhood $V \subseteq A$ of $x$ such that $\bar{V} \subseteq A$ is compact. Then $\bar{V}$ is a compact space containing the open neighborhood $V$ of $x$ , and so $A$ is locally compact.

[3.7.8]Theorem(Characterization of Open Subspaces of Compact Hausdorff Space)#

A space $X$ is homeomorphic to an open subspace of a compact Hausdorff space if and only if $X$ is locally compact Hausdorff.