Baby Compactification

Previously, we have seen that $\mathbb R$ can be compactified into the set $\mathbb R^* := \mathbb R \cup \{ \infty \}$ , which is topologically equivalent to the unit circle (or rather, $1$ -sphere) $S^1$ . We extended the usual topology on $\mathbb R$ with sets of the form $\mathbb R \backslash C \cup \{ \infty \}$ , where $C \subseteq \mathbb R$ is compact and Hausdorff.

Can we accomplish such a compactification for general topological spaces $(K, \mathcal T)$ ? Can we append a point $\infty$ to $K$ , forming $K^* := K \cup \{ \infty \}$ , and equip this set with the following topology?

$\mathcal T' := \mathcal T \cup \{ K^* \backslash C : C \subseteq K\ \text{compact}\}$

Well certainly, but the real question is whether $K^*$ is a compact Hausdorff space or not.

Let’s suppose $K^*$ is compact and Hausdorff. It is then not hard to verify that $K$ is Hausdorff, if the subspace topology agrees with the topology on $K$ .

Lemma 1. If the subspace topology of $K \subseteq K^*$ agrees with the topology on $K$ , then $K$ is Hausdorff.

How does compactness come into play? Well, if $K$ is compact then we don’t need to compactify anything, so clearly it’s only meaningful to discuss the case when $K$ is not compact. Yet, ideally, $K$ should be somewhat compact in order to be compactifiable.

Lemma 2. For any $x \in K$ , there exists a compact set $C \subseteq K$ that contains some neighbourhood $U$ of $x$ . In this case, we say that $K$ is locally compact.

Proof. Fix $x \in K$ and any neighbourhood $U$ of $x$ . Since $\bar U$ is a closed subset of the compact space $K^*$ , $\bar U$ is the required compact set.

Thus, if $K^*$ is a one-point compactification of $K$ that contains $K$ , then $K$ is Hausdorff and locally compact, but not compact. It turns out that these conditions are not just necessary, but sufficient for $K^*$ to be a one-point compactification of $K$ .

Theorem 1. $K$ is Hausdorff, locally compact, and not compact if and only if $K^* = K \cup \{ \infty \}$ is a compact Hausdorff space that contains $K$ as a subspace such that $\bar K = K^*$ .

Proof. For the subspace claim, we claim that

$\mathcal T = \{V \cap K : V \in \mathcal T'\},$

and it suffices to verify $(\supseteq )$ under the assumption $V \notin \mathcal T$ . Since $V \in \mathcal T' \backslash \mathcal T$ , $V = K^* \backslash C$ for some compact $C \subseteq K$ . Since $K$ is Hausdorff, $C$ is closed. Since $\infty \notin K$ ,

$V \cap K = (K^* \backslash \{ \infty \}) \backslash C = K \backslash C \in \mathcal T,$

since $C$ is closed and thus $K \backslash C$ is open.

For the Hausdorff claim, fix distinct $x, y \in K^*$ , so that $x \in K$ and $y = \infty$ without loss of generality. Since $K$ is locally compact, find a compact set $C \subseteq K$ that contains some neighbourhood $U \subseteq C$ of $x$ . Since $y \notin C$ , we have $y \in K \backslash C$ . Since $U \cap K \backslash C = \emptyset$ , we have that $K^*$ is Hausdorff.

Finally, for the compactness claim, let $\mathcal U = \{ U_x : x \in K \}$ be an open cover of $K^*$ . Let $U_\infty := K^* \backslash C$ for some compact $C \subseteq K$ . For each $x \in C$ , by the subspace claim we can assume $U_x \subseteq K$ without loss of generality. Then $\{ U_x : x \in C \}$ forms an open cover for $C$ . Since $C$ is compact, we can extract a finite sub-cover $\{U_{x_1},\dots,U_{x_n}\}$ . Then $\{U_{x_1},\dots,U_{x_n}, U_\infty \}$ forms a finite sub-cover of $K^*$ , yielding compactness.

The preceding arguments prove $(\Rightarrow)$ , while Lemmas 1 and 2 yield the direction $(\Leftarrow)$ .

Now suppose that $K$ is Hausdorff, locally compact, and not compact. Is $K^*$ the only way to create a compactification of $K$ ? What if there exists another compactification $L$ of $K$ ? Here, we mean that $L \backslash K$ is a single point (in the context $L = K^*$ we called this point $\infty$ ), contains $K$ as a subspace, and is compact and Hausdorff. It turns out that $L \cong K^*$ , i.e. $L$ and $K^*$ are homeomorphic to each other.

Theorem 2. $L \cong K^*$ .

Proof. Denote $\{ \infty' \} = L \backslash K$ . Check via bookkeeping that the map $f : K^* \to L$ defined by $f|_K = \mathrm{id}_K$ and $f(\infty) = \infty'$ is a homemorphism.

Definition 1. A compactification of a Hausdorff space $K$ is a topological space $L$ that is compact Hausdorff and contains $K$ as a dense subspace, i.e. $\bar K = L$ . Let $K$ be Hausdorff, locally compact, and not compact. Define $K^*$ to be the one-point compactification of $K$ .

Hence, without loss of generality, we can call $K^*$ the one point compactification of $K$ . It is, in a sense, the smallest compactification of $K$ , since we add at most one point to $K$ . A natural question to ask would be: what is the largest compactification we could impose onto $K$ ? The answer lies in the Stone–Čech compactification of $K$ .

But first, we need to take a detour and explore the Tychonoff theorem, which will come in handy when constructing Stone-Čech. What is this theorem, you might ask? It’s a simple question. We know that $[0, 1] \subseteq \mathbb R$ is compact. Is it true that $[0, 1]^J \equiv \prod_{j \in J} [0, 1]$ is compact, for even uncountable index sets $J$ ? Under the product topology, the Tychonoff theorem remarkably answers in the affirmative! We will explore this idea in the next post.

—Joel Kindiak, 2 May 25, 1739H

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Compactifying the Reals – KindiakMath

March 21, 2026 at 3:27 pm

[…] Now, in this example, we explicitly took advantage of the Heine-Borel theorem, and therefore will not be able to immediately generalise to other topological spaces. But perhaps, this example will instruct us on the key strategy to do just that, in the next post. […]

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Baby Compactification

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