Hausdorff’s Relatives

Previously, we wanted to hunt down the conditions required for a topological space to be metrizable. Some necessary conditions be that it is first-countable and Hausdorff. Their converses are certainly not immediately true. However, the cousin of first-countability is second-countability, which does in fact imply first-countability. Likewise, the property of Hausdorffness has its cousins as well.

Let’s first explore which separations Hausdorff implies. Let $K$ be a Hausdorff space.

Lemma 1. For any two distinct points $x,y \in K$ , there exists a neighbourhood $U_x$ of $x$ that does not contain $y$ . In symbols, we call any topological space satisfying this property a $\mathrm T_1$ -space.

Proof. Fix distinct $x, y \in K$ . Since $K$ is Hausdorff, there exist disjoint neighbourhoods $U_x,U_y$ of $x,y$ respectively. In particular, $y \in U_y \subseteq K \backslash U_x$ so that $y \notin U_x$ .

In fact, the Hausdorff property is often denoted by the symbol $\mathrm T_2$ , so that we yield the topological implication $\mathrm T_2 \Rightarrow \mathrm T_1$ .

Lemma 2. $K$ is $\mathrm T_1$ if and only if for any $x \in K$ , $\{ x \}$ is closed.

Proof. For the direction $(\Rightarrow)$ , suppose $K$ is $\mathrm T_1$ . Fix $x \in K$ . For any $y \in K \backslash \{x\}$ , there exists a neighbourhood $U_y$ of $y$ that does not contain $x$ . Hence, $y \in U_y \subseteq K \backslash \{x\}$ . Thus, $K \backslash \{x\}$ is open, and $\{x\}$ is closed.

For the direction $(\Leftarrow)$ , suppose $\{x\}$ is closed for any $x \in K$ . Then for any $y \in K \backslash \{x\}$ , $K \backslash \{x\}$ is a neighbourhood of $y$ that does not contain $x$ , as required.

To prove that $(\mathbb R, \mathcal T)$ is not $\mathrm T_2$ , consider the elements $0, 1 \in \mathbb R$ , and let $U_0, U_1$ be neighbourhoods of $0, 1$ respectively. Then there exist finite sets $K_0, K_1$ such that $U_0 = \mathbb R \backslash K_0$ and $U_1 = \mathbb R \backslash K_1$ . By de Morgan’s laws,

$U_0 \cap U_1 = \mathbb R \backslash (K_1 \cup K_2).$

If the left-hand side is empty, then so is the right-hand side, which means that $\mathbb R \subseteq K_1 \cup K_2$ , so that $\mathbb R$ is finite, a blatant contradiction.

To check that $\mathrm T_1 \not\Rightarrow \mathrm T_2$ in general, we use a counterexample.

Example 1. $\mathbb R$ equipped with the cofinite topology $\mathcal T$ is $\mathrm T_1$ but not $\mathrm T_2$ .

Proof. Fix distinct points $x, y \in \mathbb R$ . Then the neighbourhood $\mathbb R \backslash \{y\}$ contains $x$ but not $y$ . Since this holds for any $x, y \in \mathbb R$ , $(\mathbb R, \mathcal T)$ is $\mathrm T_1$ .

From now on, we will assume that $K$ is minimally $\mathrm T_1$ .

What, then, could be $\mathrm T_3$ ? Whatever it is, we would like it to be strong enough so that $\mathrm T_3 \Rightarrow \mathrm T_2$ and that $\mathrm T_2 \not\Rightarrow T_3$ in general. To discover new ideas, it is usually worth considering the finite case. And in topology, a sufficiently useful “finite”-ish property is compactness.

Now let’s assume $K$ is compact and $\mathrm T_2$ . Consider any closed subset $C \subseteq K$ . Fix $x \in K \backslash C$ . Since $K$ is Hausdorff, for any $y \in C$ , there exist disjoint neighbourhoods $U_y$ of $x$ and $V_y$ of $y$ . Since $C$ is closed, $K \backslash C$ is open, so that $\{V_y : y \in C\} \cup \{K \backslash C\}$ forms an open cover for $K$ . Since $x \notin V_y$ for any $y$ , a finite sub-cover for $K$ must take the form $\{V_{y_1}, \dots , V_{y_n}, K \backslash C\}$ . Define $U := \bigcap_{i=1}^n U_{y_i}$ , which contains $x$ , and $V := \bigcup_{i=1}^n V_{y_i}$ , which contains $V$ . Then

$\begin{aligned}U \cap V &= \bigcap_{i=1}^n U_{y_i} \cap \bigcup_{j=1}^n V_{y_j} \\ &= \bigcup_{j=1}^n \left( \bigcap_{i=1}^n (U_{y_i} \cap V_{y_j}) \right) \\ &\subseteq \bigcup_{j=1}^n (U_{y_j} \cap V_{y_j}) = \emptyset. \end{aligned}$

This result gives us Lemma 3.

Lemma 3. Suppose $K$ is compact and $\mathrm T_2$ . Then for any closed set $C$ and $x \in K \backslash C$ , there exist disjoint neighbourhoods $U_x$ of $x$ and $V_C$ of $C$ . We call spaces that satisfy this property regular, and denote them by $\mathrm T_3$ .

Lemma 4. For $\mathrm T_1$ spaces, $\mathrm T_3 \Rightarrow \mathrm T_2$ .

Proof. Let $K$ be regular. Fix distinct points $x, y$ . Since $K$ is $\mathrm T_1$ , $\{y\}$ is closed. Since $K$ is regular, there exist disjoint neighbourhoods $U_x$ of $x$ and $V_y$ of $\{y\}$ , the latter of which contains $y$ . Hence, $K$ is Hausdorff.

Example 2. Let $\mathbb R = (\mathbb R, \mathcal T)$ denote the usual topology on $\mathbb R$ . Define $\mathbb R_K := (\mathbb R, \mathcal S)$ by the topology $\mathcal S$ generated by basis elements of the form $(a, b) \backslash K$ , where $K := \{1/n : n \in \mathbb N\}$ . Then $\mathbb R_K$ is Hausdorff but not regular.

Proof. We leave it as an exercise to verify that $\mathbb R_K$ is Hausdorff. For non-regularity, consider the element $0$ and the closed set $K$ .

From now on, we assume that $K$ is Hausdorff.

The compact Hausdorff property strengthened the $\mathrm T_2$ property into the $\mathrm T_3$ property, which implies that $\mathrm T_2$ property, assuming that $\mathrm T_1$ holds. What else can compact Hausdorff spaces do for us? We could separate a point from a closed set by disjoint neighbourhoods—that’s the $\mathrm T_3$ property. Could we separate two closed sets by disjoint neighbourhoods?

Lemma 4. Suppose $K$ is compact and $\mathrm T_3$ . Then for disjoint closed subsets $B, C \subseteq K$ , there exist disjoint neighbourhoods $U$ of $B$ and $V$ of $C$ . In this case, we call the space normal, denoted $\mathrm T_4$ .

Proof. Let $B, C$ be disjoint closed subsets of $K$ . Since $K$ is regular, for any $x \in B$ , there exist disjoint neighbourhoods $U_x$ of $x$ and $V_x$ of $C$ . Then

$\{U_x : x \in B\} \cup \{K \backslash B\}$

forms an open cover for $K$ . Since $K$ is compact, extract a finite sub-cover

$\{U_{x_1},\dots, U_{x_n}, K \backslash B\}$

of $K$ . Define the neighbourhoods $U := \bigcup_{i=1}^n U_{x_i}$ of $B$ and $V := \bigcap_{j=1}^n V_{x_j}$ of $C$ . By the argument in Lemma 3, $U$ and $V$ are disjoint, as required.

Lemma 5. Let $K$ be a Hausdorff space.

$K$ is regular if and only if for any $x \in K$ and neighbourhood $U$ of $x$ , there exists a neighbourhood $V$ of $x$ such that $\bar{V} \subseteq U$ .
$K$ is normal if and only if for any closed $C \subseteq K$ and neighbourhood $U$ of $C$ , there exists a neighbourhood $V$ of $C$ such that $\bar{V} \subseteq U$ .

Proof. We prove the case for normality, and the case for regularity follows by setting $C = \{x\}$ . For the direction $(\Rightarrow)$ , fix the closed set $C \subseteq K$ and a neighborhood $U \supseteq C$ . Then $K \backslash U$ and $C$ are disjoint closed sets. Using normality, find disjoint neighbourhoods $W$ of $K \backslash U$ and $V$ of $C$ . Since $K \backslash W$ is closed and contains $V$ ,

$C \subseteq V \subseteq \bar V \subseteq K \backslash W \subseteq U,$

as required. For the direction $(\Leftarrow)$ , fix disjoint closed subsets $B, C$ of $K$ . Then $K \backslash C$ is a neighbourhood of $B$ . Find a neighbourhood $U$ of $K \backslash C$ such that $\bar U \subseteq K \backslash C$ . Then $V := K \backslash \bar U$ is a neighbourhood of $C$ disjoint from $U$ , as required.

Corollary 1. For Hausdorff spaces, $\mathrm T_4 \Rightarrow \mathrm T_3$ .

To prove that $\mathrm T_3 \not\Rightarrow \mathrm T_4$ takes a nontrivial amount of effort, and we shall omit that undertaking until absolutely necessary.

Now, in the grand scheme of things, who cares about normal spaces? It turns out that normality coupled with second-countability gives us the ticket to creating metrizable spaces. But we will require some nontrivial effort to summit that peak.

—Joel Kindiak, 23 Apr, 25, 1659H

KindiakMath

Hausdorff’s Relatives

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Hausdorff’s Relatives

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