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 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

KindiakMath

Baby Compactification

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Baby Compactification

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