Regular (T3) and Normal (T4) Spaces

Regularity and normality extend the separation axioms beyond point separation to the separation of points from closed sets and closed sets from each other. These stronger conditions enable powerful tools such as the Urysohn lemma and the Tietze extension theorem.

Regular Spaces

Definition6.4Regular Space

A topological space $X$ is regular if for every point $x \in X$ and every closed set $C$ not containing $x$ , there exist disjoint open sets $U$ and $V$ with $x \in U$ and $C \subseteq V$ .

A space is $T_3$ if it is regular and $T_1$ (equivalently, regular and $T_0$ , since regular + $T_0$ implies $T_1$ ).

Theorem6.3Characterization of Regularity

A $T_1$ space $X$ is regular if and only if for every $x \in X$ and every open neighborhood $U$ of $x$ , there exists an open $V$ with $x \in V \subseteq \overline{V} \subseteq U$ .

Proof

( $\Rightarrow$ ): Let $U$ be open with $x \in U$ . Then $C = X \setminus U$ is closed and $x \notin C$ . By regularity, there exist disjoint open $V \ni x$ and $W \supseteq C$ . Then $V \cap W = \emptyset$ implies $V \subseteq X \setminus W$ . Since $X \setminus W$ is closed, $\overline{V} \subseteq X \setminus W \subseteq X \setminus C = U$ .

( $\Leftarrow$ ): Given $x \notin C$ (closed), set $U = X \setminus C$ . By hypothesis, there exists open $V$ with $x \in V \subseteq \overline{V} \subseteq U$ . Then $V$ and $X \setminus \overline{V}$ are disjoint open sets separating $x$ from $C$ .

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ExampleRegular and Non-Regular Spaces

Every metric space is regular: Given $x$ and a closed $C$ with $x \notin C$ , set $d = d(x, C) > 0$ . Then $B(x, d/2)$ and $\{y : d(y, C) < d/2\}$ are disjoint open neighborhoods.
$\mathbb{R}$ with the $K$ -topology ( $\mathcal{B}_K = \{(a,b)\} \cup \{(a,b) \setminus K\}$ where $K = \{1/n\}$ ) is Hausdorff but not regular: the closed set $K$ cannot be separated from $0$ by disjoint open sets.
The Sorgenfrey line $\mathbb{R}_\ell$ is regular (and in fact normal).

Normal Spaces

Definition6.5Normal Space

A topological space $X$ is normal if for every pair of disjoint closed sets $C, D$ , there exist disjoint open sets $U, V$ with $C \subseteq U$ and $D \subseteq V$ .

A space is $T_4$ if it is normal and $T_1$ .

Theorem6.4Characterization of Normality

A $T_1$ space $X$ is normal if and only if for every closed set $C$ and open set $U$ with $C \subseteq U$ , there exists an open $V$ with $C \subseteq V \subseteq \overline{V} \subseteq U$ .

Proof

( $\Rightarrow$ ): Given $C \subseteq U$ with $C$ closed and $U$ open, set $D = X \setminus U$ . Then $C$ and $D$ are disjoint closed sets. By normality, there exist disjoint open $V \supseteq C$ and $W \supseteq D$ . Then $\overline{V} \cap W = \emptyset$ (since $V \cap W = \emptyset$ and $W$ is open), so $\overline{V} \subseteq X \setminus W \subseteq X \setminus D = U$ .

( $\Leftarrow$ ): For disjoint closed sets $C, D$ , take $U = X \setminus D$ . By hypothesis, $C \subseteq V \subseteq \overline{V} \subseteq U$ . Then $V$ and $X \setminus \overline{V}$ are the desired disjoint open sets.

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Examples and Counterexamples

ExampleNormal Spaces

Every metric space is normal (a consequence of the Urysohn metrization approach, or proved directly using distance functions).
Every compact Hausdorff space is normal (Theorem 5.10).
Every second-countable regular space is normal (consequence of the Urysohn metrization theorem).
$\mathbb{R}$ and $\mathbb{R}^n$ with the standard topology are normal.
The Sorgenfrey line $\mathbb{R}_\ell$ is normal.

ExampleNon-Normal Spaces

The Sorgenfrey plane $\mathbb{R}_\ell \times \mathbb{R}_\ell$ is not normal, even though each factor is normal. This shows normality is not preserved by products.
$\mathbb{R}^\omega$ in the box topology is not normal (Hausdorff but not normal).
The bug-eyed line (two copies of $\mathbb{R}$ identified except at the origin) is Hausdorff but not regular (hence not normal).

The Separation Axiom Hierarchy

RemarkThe Complete Hierarchy

$T_4 \; (\text{normal} + T_1) \implies T_3 \; (\text{regular} + T_1) \implies T_2 \; (\text{Hausdorff}) \implies T_1 \implies T_0$

Each implication is strict. Additionally:

Normal does not imply regular without $T_1$ (pathological counterexamples exist).
$T_4$ does not imply metrizable (the Sorgenfrey line is $T_4$ but not metrizable).
All metric spaces are $T_4$ , but $T_4$ is not preserved by products.

The hierarchy continues with completely regular ( $T_{3\frac{1}{2}}$ ) spaces, which lie between $T_3$ and $T_4$ in terms of their relationship to function separation.

Theorem6.5Normal Implies Regular for $T_1$ Spaces

Every $T_4$ space is $T_3$ .

Proof

Let $X$ be $T_4$ and let $x \in X$ with $C$ closed and $x \notin C$ . Since $X$ is $T_1$ , $\{x\}$ is closed. Then $\{x\}$ and $C$ are disjoint closed sets. By normality, there exist disjoint open sets $U \ni x$ and $V \supseteq C$ . So $X$ is regular.

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