Chapter 13: Separation Axioms

In point-set topology, separation axioms are used to characterize the degree to which points, points and closed sets, and closed sets can be separated by open sets in a topological space. In the previous lecture we discussed the hierarchical relationship of the separation axioms:

\[ T_0 < T_1 < T_2 < T_3 < T_4 \]

In this chapter we will delve into the equivalent characterizations and important properties of these axioms.

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1. \(T_1\) spaces and \(T_2\) spaces (Hausdorff)

Lemma 13.1 (Characterization of \(T_1\) spaces by singleton sets)

For a topological space \((X, \mathcal{T})\), the following conditions are equivalent:

• (1) \(X\) is a \(T_1\) space.

• (2) For every \(x \in X\), the singleton set \(\{x\}\) is closed.

Examples analysis:

• (1) Every metric space \((X, d)\) is a \(T_2\) space (and hence also \(T_1\)).

• (2) Let \(X\) be a set containing infinitely many points (i.e., \(\text{Card}(X) = \infty\)). Equip it with the cofinite topology:

\[ \tau = \{\emptyset, X\} \cup \{X - S \mid S \text{ is a finite set}\} \]

This space is \(T_1\), because the complement of any singleton set is a finite set and thus open, so every singleton is closed. However, it is not \(T_2\), because in this topology any two nonempty open sets have nonempty intersection, so it is impossible to separate two distinct points by disjoint open sets.

Lemma 13.2 (Diagonal characterization of \(T_2\) spaces)

For a topological space \((X, \mathcal{T})\), the following conditions are equivalent:

• (1) \(X\) is a \(T_2\) space (i.e., Hausdorff space).

• (2) The diagonal set \(\Delta = \{(x,x) \mid x \in X\}\) is closed in the product space \(X \times X\).

Proof of Lemma 13.2

(1) \(\Rightarrow\) (2):

We want to show that \(X \times X - \Delta\) is open.

Take an arbitrary \((x,y) \in X \times X - \Delta\). This means \(x \ne y\).

Since \(X\) is \(T_2\), there exist open sets \(U_x\) and \(U_y\) containing \(x\) and \(y\) respectively such that:

\[ U_x \cap U_y = \emptyset \]

This implies that the open set \(U_x \times U_y\) in the product space does not intersect the diagonal:

\[ \Delta \cap (U_x \times U_y) = \emptyset \]

Therefore the open neighborhood \(U_x \times U_y\) satisfies:

\[ U_x \times U_y \subseteq X \times X - \Delta \]

Hence \(X \times X - \Delta\) is open, i.e., \(\Delta\) is closed.

(2) \(\Rightarrow\) (1):

Simply reverse the reasoning above. If \(\Delta\) is closed, its complement is open. For any point \((x,y)\) in the complement, we can find a basis element \(U_x \times U_y\) that does not intersect the diagonal, which implies \(U_x\) and \(U_y\) are disjoint, satisfying the \(T_2\) property. \(\square\)

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2. Regular spaces and \(T_3\) spaces

Definition 13.1 (Regular Space)

A topological space \((X, \mathcal{T})\) is called regular if for every closed set \(A \subseteq X\) and every point \(x \in X - A\) (i.e., \(x \notin A\)), there exist open sets \(U_A\) and \(U_x\) such that:

\[ A \subseteq U_A, \quad x \in U_x, \quad U_A \cap U_x = \emptyset \]

If a space is both regular and \(T_1\) (hence singleton sets are closed), it is called a \(T_3\) space.

Remarks:

• (1) \(T_3 \implies T_2\). Because with the \(T_1\) property, singletons are closed; applying regularity to separate a singleton and a point not belonging to it is exactly the \(T_2\) separation condition.

• (2) The terminology here differs from some textbooks (e.g., Munkres). In some references, regularity already includes \(T_1\); please follow our definitions which clearly distinguish between regular and \(T_3\).

Classical counterexample: The \(K\)-topology

Let \(A = \{1/n \mid n \ge 1\}\). Consider the topology \(\tau_K\) on \(\mathbb{R}\) generated by the following basis:

\[ \mathcal{B} := \{(a,b) \mid a, b \in \mathbb{R}\} \cup \{(a,b) - A \mid a, b \in \mathbb{R}\} \]

The space \((\mathbb{R}, \tau_K)\) is \(T_2\), but it is not \(T_3\). The reason is that the set \(A\) is closed in this topology and \(0 \notin A\). However, any open set containing \(0\) must intersect any open set containing \(A\), so they cannot be separated.

Lemma 13.3 (Equivalent characterization of regularity)

For a topological space \((X, \mathcal{T})\), the following conditions are equivalent:

• (1) \(X\) is regular.

• (2) For every point \(x\) and every open neighborhood \(U\) of \(x\) (i.e., \(x \in U\)), there exists an open set \(V\) such that:

\[ x \in V \subseteq \overline{V} \subseteq U \]

Proof of Lemma 13.3

(1) \(\Rightarrow\) (2):

Suppose \(x \in U\) with \(U\) open. Consider the closed set \(X - U\). Since the space is regular, the point \(x\) and the closed set \(X-U\) can be separated.

There exist open sets \(V \ni x\) and \(W \supseteq X - U\) such that:

\[ V \cap W = \emptyset \]

This implies \(V \subseteq X - W\). Since \(X - W\) is closed (it is the complement of the open set \(W\)), the smallest closed set containing \(V\), namely \(\overline{V}\), must also be contained in \(X - W\):

\[ \overline{V} \subseteq X - W \subseteq X - (X - U) = U \]

Therefore, there exists an open set \(V\) such that \(x \in V \subseteq \overline{V} \subseteq U\).

(2) \(\Rightarrow\) (1):

Given a closed set \(A\) and a point \(x \in X - A\). Since \(X - A\) is open, by condition (2) there exists an open set \(V\) such that:

\[ x \in V \subseteq \overline{V} \subseteq X - A \]

Set \(U_A = X - \overline{V}\). Since \(\overline{V}\) is closed, \(U_A\) is open and clearly \(A \subseteq U_A\).

Moreover, we have:

\[ V \cap U_A = V \cap (X - \overline{V}) = \emptyset \]

Thus \(x\) and \(A\) are separated by disjoint open sets \(V\) and \(U_A\), proving the space is regular. \(\square\)

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3. Normal spaces and \(T_4\) spaces

Definition 13.2 (Normal Space)

A topological space \((X, \mathcal{T})\) is called normal if for any two disjoint closed sets \(A, B \subseteq X\) (i.e., \(A \cap B = \emptyset\)), there exist open sets \(U, V\) such that:

\[ A \subseteq U, \quad B \subseteq V, \quad U \cap V = \emptyset \]

If a space is both normal and \(T_1\), it is called a \(T_4\) space.

Remarks:

• (1) \(T_4 \implies T_3 \implies T_2 \implies T_1 \implies T_0\).

• (2) Without the \(T_1\) hypothesis, a normal space is not necessarily regular.

• (3) Every metric space \((X, d)\) is a \(T_4\) space.

Similar to regularity, normality also has an equivalent characterization via closures:

Lemma 13.4 (Equivalent characterization of normality)

For a topological space \((X, \mathcal{T})\), the following conditions are equivalent:

• (1) \(X\) is normal.

• (2) Given any closed set \(A\) and an open neighborhood \(U\) of \(A\) (i.e., \(A \subseteq U\)), there exists an open set \(V\) such that:

\[ A \subseteq V \subseteq \overline{V} \subseteq U \]

3.1 Subspace and product properties of separation axioms

• Subspace heredity: Every subspace of a \(T_2\) (or \(T_3\)) space is again \(T_2\) (or \(T_3\)). However, normality is generally not hereditary under subspaces.

• Product space property: An arbitrary product space \(\prod_{i \in I} X_i\) is \(T_2\) (or \(T_3\)) if and only if each coordinate space \(X_i\) is \(T_2\) (or \(T_3\)).

• Compactness boost: A space that is both \(T_2\) (Hausdorff) and compact is necessarily \(T_4\).

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4. Regularity plus second countability implies normality

This is an extremely important theorem that serves as a bridge in the proof of the Urysohn metrization theorem.

Proposition 13.5

A topological space that is regular and second countable is necessarily normal.

Therefore, \(T_3\) + second countable \(\implies T_4\).

Proof of Proposition 13.5

Let \((X, \mathcal{T})\) be a regular and second countable space, and let \(\mathcal{B} = \{B_i\}_{i \in \mathbb{N}}\) be a countable basis for its topology.

Given any two disjoint closed sets \(A, C \subseteq X\).

First, for a point \(x \in A\), since \(x \notin C\) and the space is regular, by Lemma 13.3 there exists a basis element \(U_j \in \mathcal{B}\) such that:

\[ x \in U_j \subseteq \overline{U_j} \subseteq X - C \]

Let \(x\) run through \(A\); we pick from \(\mathcal{B}\) a family of open sets that covers \(A\). Because the basis is countable, the family we pick is also countable. We rearrange it as a sequence \(\{U_n\}_{n \ge 1}\). This sequence satisfies:

\[ A \subseteq \bigcup_{n \ge 1} U_n, \quad \text{and} \quad \overline{U_n} \subseteq X - C \]

Similarly, for the closed set \(C\), we pick from \(\mathcal{B}\) a countable sequence \(\{V_n\}_{n \ge 1}\) that covers \(C\) and satisfies:

\[ C \subseteq \bigcup_{n \ge 1} V_n, \quad \text{and} \quad \overline{V_n} \subseteq X - A \]

The challenge now is that although these two countable open covers each avoid the other closed set, they may intersect each other (i.e., some \(U_j\) may intersect some \(V_k\)). We need to systematically "trim" them to obtain two completely disjoint open sets.

For each \(n \ge 1\), define new open sets:

\[ U_n' = U_n - \bigcup_{i=1}^n \overline{V_i} \]

\[ V_n' = V_n - \bigcup_{i=1}^n \overline{U_i} \]

Since each subtraction is of a finite union of closed sets (which is still closed), \(U_n'\) and \(V_n'\) are open.

Now define the overall covering open sets:

\[ U' := \bigcup_{n \ge 1} U_n', \quad V' := \bigcup_{n \ge 1} V_n' \]

Claim 1: \(A \subseteq U'\) and \(C \subseteq V'\)

For any \(x \in A\), there exists a first \(U_n\) that covers it. Because \(A \cap \overline{V_i} = \emptyset\) (all \(\overline{V_i}\) are contained in \(X - A\)), subtracting any \(\overline{V_i}\) from \(U_n\) does not remove \(x\). Hence \(x \in U_n' \subseteq U'\). The same argument holds for \(C\).

Claim 2: \(U' \cap V' = \emptyset\)

Suppose there exists a point \(z \in U' \cap V'\). Then there must exist indices \(j\) and \(k\) such that:

\[ z \in U_j' \quad \text{and} \quad z \in V_k' \]

Without loss of generality, assume \(j \le k\) (by symmetry).

By the definition of \(V_k'\), we have:

\[ V_k' = V_k - \bigcup_{i=1}^k \overline{U_i} \]

Since \(j \le k\), in constructing \(V_k'\) the set \(\overline{U_j}\) has been explicitly removed. This means that \(V_k'\) cannot contain any point belonging to \(\overline{U_j}\).

However, because \(z \in U_j'\) and by definition \(U_j' \subseteq U_j \subseteq \overline{U_j}\), it follows that \(z\) must belong to \(\overline{U_j}\). This is a direct contradiction!

Therefore the assumption is false, and \(U'\) and \(V'\) are disjoint. This proves that \(X\) is normal. \(\square\)

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