Compactification on Steroids

We return to answer our golden question: what’s the largest compactification we can obtain on a sufficiently reasonable topological space? We’ll need the nuclear weapons of Urysohn’s lemma and the Tychonoff theorem to answer this question.

Now, Urysohn’s lemma tells us that if $K$ is a normal Hausdorff topological space, then for disjoint closed subsets $C, D$ of $K$ , we can find a continuous function $f : K \to [0, 1]$ with the special feature that $f(C) = \{0 \}$ and $f(D) = \{1\}$ . This theorem certainly still holds in the particular case $D = \{x\}$ for any $x \notin C$ . We call such a space completely regular, and formally define it below.

Definition 1. A Hausdorff space $K$ is completely regular if for any closed set $C$ and $x \in K \backslash C$ , there exists a continuous function $f : K \to [0, 1]$ such that $f(C) = \{0\}$ and $f(x) = 1$ . As per previous discussions, normal spaces are completely regular.

Furthermore, completely regular spaces are regular, so that our naming of such a space is at least consistent with prior ideas of regular spaces.

Lemma 1. Completely regular spaces are regular.

Let’s also recall the imbedding theorem, that was ironically enough embedded in the proof of the Urysohn metrization theorem.

Lemma 2 (Imbedding Theorem). Suppose $K$ is Hausdorff. Let $\{f_\alpha\}_{\alpha \in J}$ be a collection of functions $f_\alpha : K \to [0, 1]$ , where for any $x \in K$ and neighbourhood $U$ of $x$ , there exists $\alpha \in J$ such that $f_\alpha (x) > 0$ and $f(K \backslash U) = \{0 \}$ . Then the function $f : K \to [0, 1]^J$ defined by

$f(x) := (f_\alpha(x))_{\alpha \in J}$

imbeds $K$ into $[0, 1]^J$ (i.e. $f : K \to f(K)$ is a homeomorphism).

Lemma 3. If $K$ is metrizable, then it is normal, and thus completely regular.

Proof. Let $d$ be a metric on $K$ and $C, D$ be disjoint closed subsets of $K$ . Now, for any $x \in K$ , the set $\{d(x, y) : y \in C\}$ is bounded below and has a non-negative infimum, which we will denote $d(x, C)$ .

We observe that if $x \notin C \!\! \iff \!\! x \in K \backslash C$ , then there exists sufficiently small $\epsilon > 0$ such that $B_d(x, \epsilon) \subseteq K \backslash C$ . Hence, for any $y \in C$ , $d(x, y) \geq \epsilon$ . Therefore, $d(x, C) \geq \epsilon > 0$ . By a similar argument we can prove the reverse direction, yielding $d(x, C) > 0$ if and only if $x \notin C$ . Hence, $d(C,C) = 0$ .

By the reverse triangle inequality, it is not hard to verify that the map $d( \cdot, C) : K \to \mathbb R$ is continuous. Finally, define the continuous map $f : K \to \mathbb R$ by

$\displaystyle f(x) := \frac{d(x, C)}{d(x, C) + d(x, D)},$

which is well-defined since the denominator is guaranteed to be nonzero. Since $f(C) = \{0 \}$ and $f(D) = \{1\}$ , $C \subseteq f^{-1}((-\infty,1/3))$ and $D \subseteq f^{-1}((2/3, \infty))$ , so that $K$ is normal, as required.

Lemma 4. $K$ is completely regular if and only if it is homeomorphic to a subspace of $[0, 1]^J$ for some $J$ .

Proof. For the direction $(\Rightarrow)$ , use completely regularity to define, for any $x \in K$ and neighbourhood $U_x$ of $x$ , the continuous function $f_x : K \to [0, 1]$ such that $f_x(x) > 0$ and $f_x(K \backslash U_x) = \{ 0 \}$ . Then $\{f_x\}_{x \in K}$ forms the desired collection in Lemma 2. For the direction $(\Leftarrow)$ , if $f : K \to [0, 1]^J$ denotes a given imbedding, then $K \cong f(K) \subseteq [0, 1]^J$ is metrizable and thus completely regular by Lemma 3.

Completely regular spaces are nicer to work with than normal spaces since they are preserved under “usual” combinations:

Lemma 5. Subspaces and finite products of of completely regular spaces are completely regular.

It is in the completely regular setting that we want to talk about compactifications. In the one-point compactification discussion, we assumed that the topological spaces are locally compact and Hausdorff. We verify that they are completely regular too, and thus will have some kind of “largest” compactifications.

Lemma 6. Locally compact Hausdorff spaces are completely regular.

Proof. Let $K$ be a locally compact Hausdorff space with one-point compactification denoted by $\alpha K$ . Since the latter is compact Hausdorff, it is normal and thus completely regular. Since $K \subseteq \alpha K$ , it is completely regular too.

It turns out that completely regular spaces have largest compactifications, which we need to prove both the (i) existence, and (ii) uniqueness, at least up to homeomorphisms. The key ingredient we need will be the compact Hausdorff space.

Lemma 7. Suppose $M$ is a compact Hausdorff space and $f : K \to M$ is an imbedding. Then there exists a compactification $L \supseteq K$ that is associated to an imbedding $F : L \to M$ such that $F|_K = f$ . Any two compactifications $L_1, L_2 \supseteq K$ with said property are equivalent in the following sense: there exists a homeomorphism $G : L_1 \to L_2$ such that $G|_K = \mathrm{id}_K$ .

Proof. Since $f$ is an imbedding, $K \cong f(K)$ . Then $\overline{f(K)} \subseteq M$ is compact Hausdorff, i.e. a compactification of $f(K)$ . Intuitively, we want to create the topological space $L \supseteq K$ satisfying $L \cong \overline{f(K)}$ , so that $L$ is a compactification of $K$ .

To that end, we take advantage of the set $H := \overline{f(K)} \backslash f(K)$ . If we identify $x \longleftrightarrow f(x)$ , then we obtain $H \cap K = \emptyset$ . Hence, we define $L := H \cup K$ and the map $F : L \to M$ by

$F|_H = \mathrm{id}_H,\quad F|_K = f.$

Define the topology $\mathcal T$ on $L$ by

$\mathcal T := \{ U \subseteq L : F(U) \subseteq \overline{f(K)} \subseteq M\ \text{open}\}.$

Then $F : L \to M$ is the desired homeomorphism. For uniqueness, let $L_i$ be compactifications of $K$ and $F_i : L_i \to M$ be homeomorphisms such that $F_i|_K = f$ . Then $F_2^{-1} \circ F_1 : L_1 \to L_2$ is the desired homemorphism.

Lemma 7 almost gets us to our desired result. Notice that for Lemma 7, given an imbedding $f : K \to M$ , we can find a corresponding sufficiently unique compactification $L$ of $K$ . The Stone-Čech compactification is the “largest” such compactification in the sense of reversing the quantifiers: we want a compactification that works for any bounded continuous map $f$ .

Theorem 1. Let $K$ be a completely regular space. Then there exists a compactification $L \supseteq K$ such that any bounded continuous map $f : K \to \mathbb R$ extends uniquely to a bounded continuous map $F : L \to \mathbb R$ , where $F|_K = f$ .

Proof. By Lemma 7, it suffices to embed $K$ in a compact Hausdorff space. Let $\{f_\alpha\}$ note the collection of all bounded and continuous real-valued functions on $K$ . For each $\alpha$ , define

$I_\alpha := [\inf f_\alpha(K), \sup f_\alpha(K)].$

By the Tychonoff theorem, the product $C := \prod_\alpha I_\alpha$ is compact. Since each $I_\alpha$ is Hausdorff, so is $C$ . Define the map $g : K \to C \subseteq \prod_\alpha \mathbb R^\alpha$ by $g(x) = (f_\alpha(x))_\alpha$ . Since $K$ is completely regular, $\{f_\alpha\}$ separates points from closed sets in $K$ . Hence, the imbedding theorem yields the imbedding $g$ .

By Lemma 7, there exists a compactification $L$ of $K$ and an extension $G : L \to C$ of $g : K \to C$ such that $G|_K = g$ . Fix any bounded continuous map $f : K \to \mathbb R$ . We extend it uniquely to $F : L \to \mathbb R$ . Let $f = f_\alpha$ for some $\alpha$ . Then the desired extension is $F := \pi_\alpha \circ G : L \to \mathbb R$ , since for any $x \in K$ ,

$F(x) = (\pi_\alpha \circ G)(x) = \pi_\alpha(G(x)) = \pi_\alpha(g(x)) = f_\alpha(x) = f(x).$

For uniqueness, let $F_1,F_2 : \bar K \to \mathbb R$ be extensions of $f$ . If $F_1 \neq F_2$ , then find $x \in \bar K$ such that $F_1(x) \neq F_2(x)$ . Since $\mathbb R$ is Hausdorff, find disjoint neighbourhoods $U_1,U_2$ of $F_1(x),F_2(x)$ respectively. Find a neighbourhood $V$ of $x$ such that

$F_1(V) \subseteq U_1,\quad F_2(V) \subseteq U_2,$

contradicting $U_1 \cap U_2 = \emptyset$ .

Corollary 1. Let $L$ be the compactification of $K$ in Theorem 1 and $C$ be any compact Hausdorff space. Then any bounded continuous map $f : K \to C$ extends uniquely to a bounded continuous map $F : L \to C$ , where $F|_K = f$ .

Proof. Fix any compact Hausdorff space $C$ . Since $C$ is compact Hausdorff, it is completely regular, and thus can be imbedded into $[0, 1]^J$ . Henceforth, we will assume $C = [0, 1]^J$ . Given any $f : K \to C$ , the component function $f_\alpha : K \to [0, 1]$ is bounded and continuous. By Theorem 1, each $f_\alpha$ can be extended to some unique continuous map $F_\alpha : L \to [0, 1]$ .

Define the continuous map $F : K \to \mathbb R^J$ by $F(x) = (F_\alpha(x))_\alpha$ . Then

$F(L) = F(\bar K) \subseteq \overline{F(K)} = \overline{f(K)} \subseteq \bar C = C,$

as required.

Corollary 2. Any two compactifications $L_1, L_2$ described in Theorem 1 are equivalent to each other.

Proof. Let $i_1 : K \to L_1$ and $i_2 : K \to L_2$ denote the continuous inclusion maps into compact Hausdorff spaces. By the extension property, extend $i_1$ to $I_1 : L_1 \to L_2$ such that $I_1 |_K = i_1$ . Obtain an extension $I_2 : L_2 \to L_1$ similarly. Then $I_2 \circ I_1 : L_1 \to L_1$ is a continuous extension of $\mathrm{id}_{L_1} : L_1 \to L_1$ satisfying $(I_2 \circ I_1) \mid_{L_1} = \mathrm{id}_{L_1}$ , so that $L_1$ and $L_2$ are equivalent to each other.

Let’s take a step back for a moment:

We first observed some pleasant properties of completely regular spaces that are more powerful than regular spaces but not as restrictive as normal spaces (i.e. Lemmas 1–6).
We proved that given any continuous map $f : K \to C$ , where $C$ is compact and Hausdorff, $K$ has a compactification $L$ and $f$ has an extension $F : L \to C$ such that $F|_K = f$ (i.e. Lemma 7).
We proved the *stronger* claim that given any completely regular space $K$ , there exists a compactification $L$ such that any continuous map $f : K \to C$ , where $C$ is compact Hausdorff, extends uniquely to a continuous map $F : K \to C$ (i.e. Theorem 1, Corollary 1).
The compactification defined in Theorem 1 must be unique (Corollary 2).

Put it more concisely, Definition 2 below is well-defined.

Definition 2. Let $K$ be any completely regular space. The compactification defined in Theorem 1, denoted $\beta K$ , is called the Stone-Čech compactification of $K$ .

The final result demonstrates how $\beta K$ is, in a sense, the “largest” compactification of $K$ .

Corollary 3. Let $L$ be any compactification of $K$ (i.e. $\bar K \subseteq L$ ). Then there exists a continuous surjective map $g : \beta K \to L$ , where for any closed set $C \subseteq \beta K$ , $g(C) \subseteq L$ is closed.

Proof. Let $i : K \to L$ denote the continuous inclusion map from $K$ to $L$ . Since $L$ is compact Hausdorff, Corollary 1 allows us to extend $i$ to the continuous map $g : \beta K \to L$ . Since $\overline{i(K)} = \bar K = L$ , we have

$\overline{g(\beta K)} = \overline{g(K)} = \overline{i(K)} = L.$

Finally, for any closed set $C \subseteq \beta K$ , since the latter is compact Hausdorff, $C$ must be compact. By the extreme value theorem, $g(C) \subseteq L$ is compact, and therefore closed. In particular, setting $C = \beta K$ yields

$g(\beta K) = \overline{g(\beta K)} = \overline{i(K)} = L,$

yielding the surjectivity of $g$ .

—Joel Kindiak, 16 May 25, 1758H

KindiakMath

Compactification on Steroids

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Compactification on Steroids

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