Unit 5: Separation Axioms: T0, T1, Hausdorff (T2) & Regular Spaces (T3)

1. The Separation Hierarchy

The separation axioms (named $T$ from the German Trennungsaxiom) measure the extent to which points and sets can be distinguished by the open sets of the topology.

2. Kolmogorov Spaces ($T_0$)

Definition 5.1 ($T_0$ / Kolmogorov Axiom): A topological space $(X, \mathcal{T})$ is called a $T_0$ space (or Kolmogorov space) if for every pair of distinct points $x \ne y \in X$, there exists an open set $U \in \mathcal{T}$ that contains one of the points but not the other:

Example 5.1 (Sierpiński Space is $T_0$ but not $T_1$): On $X = \{0, 1\}$ with $\mathcal{T} = \{\emptyset, \{1\}, \{0, 1\}\}$, the open set $\{1\}$ contains $1$ but not $0$. Thus Sierpiński space is $T_0$. However, there is no open set containing $0$ without $1$ (the only open set containing $0$ is the whole space $X$). Hence Sierpiński space fails $T_1$!

3. Fréchet Spaces ($T_1$)

Definition 5.2 ($T_1$ / Fréchet Axiom): A topological space $(X, \mathcal{T})$ is called a $T_1$ space (or Fréchet space) if for every pair of distinct points $x \ne y \in X$, there exist open sets $U, V \in \mathcal{T}$ such that:

Theorem 5.1 (Characterization of $T_1$ via Singletons): A topological space $(X, \mathcal{T})$ is $T_1$ if and only if every singleton set $\{x\}$ is closed in $X$.

Proof: ($\Rightarrow$) Assume $X$ is $T_1$: Let $x \in X$. To show $\{x\}$ is closed, we prove $X \setminus \{x\}$ is open. For any $y \in X \setminus \{x\}$, $y \ne x$. Since $X$ is $T_1$, there exists an open set $V_y$ containing $y$ such that $x \notin V_y$. This means $V_y \subseteq X \setminus \{x\}$. Then $X \setminus \{x\} = \bigcup_{y \ne x} V_y$, which is a union of open sets, hence open. Thus $\{x\}$ is closed.

($\Leftarrow$) Assume all singletons are closed: Let $x \ne y \in X$. Since $\{y\}$ is closed, $U = X \setminus \{y\}$ is open. $x \in U$ and $y \notin U$. Similarly, $\{x\}$ is closed, so $V = X \setminus \{x\}$ is open, with $y \in V$ and $x \notin V$. Thus $X$ is $T_1$. $\blacksquare$

Corollary 5.1: In any $T_1$ space, every finite subset is closed.

1. Hausdorff Spaces ($T_2$)

Introduced by Felix Hausdorff in his seminal 1914 Grundzüge der Mengenlehre, the $T_2$ axiom is the cornerstone of modern topology and analysis.

Definition 5.3 ($T_2$ / Hausdorff Space): A topological space $(X, \mathcal{T})$ is called a Hausdorff space (or $T_2$ space) if for every pair of distinct points $x \ne y \in X$, there exist disjoint open neighborhoods $U, V \in \mathcal{T}$ such that:

2. Uniqueness of Sequence Limits

Theorem 5.2 (Limits are Unique in Hausdorff Spaces): Let $(X, \mathcal{T})$ be a Hausdorff space. If a sequence $(x_n)_{n=1}^\infty$ converges to $x$ and also converges to $y$, then:

Proof: Suppose for contradiction that $x \ne y$. Since $X$ is Hausdorff, there exist disjoint open sets $U, V$ with $x \in U$, $y \in V$, and $U \cap V = \emptyset$. Since $x_n \to x$, there exists $N_1$ such that $x_n \in U$ for all $n \ge N_1$. Since $x_n \to y$, there exists $N_2$ such that $x_n \in V$ for all $n \ge N_2$. For any $n \ge \max\{N_1, N_2\}$, $x_n \in U \cap V$. This contradicts $U \cap V = \emptyset$! Therefore, $x = y$. $\blacksquare$

3. The Closed Diagonal Characterization

A profound global geometric characterization of the Hausdorff property:

Theorem 5.3 (Closed Diagonal Theorem): A topological space $X$ is Hausdorff if and only if the diagonal:

is a closed subset of the product space $X \times X$ equipped with the product topology.

Proof: ($\Rightarrow$) Assume $X$ is Hausdorff: We show $(X \times X) \setminus \Delta$ is open. Let $(x, y) \in (X \times X) \setminus \Delta$. Then $x \ne y$. Since $X$ is Hausdorff, there exist open sets $U, V \subseteq X$ with $x \in U, y \in V$, and $U \cap V = \emptyset$. The product $U \times V$ is open in $X \times X$ and contains $(x, y)$. If $(z, z) \in (U \times V) \cap \Delta$, then $z \in U \cap V = \emptyset$, contradiction! Thus $(U \times V) \cap \Delta = \emptyset$, meaning $U \times V \subseteq (X \times X) \setminus \Delta$. Thus $(X \times X) \setminus \Delta$ is open, so $\Delta$ is closed.

($\Leftarrow$) Assume $\Delta$ is closed: If $x \ne y$, then $(x, y) \notin \Delta$. Since $(X \times X) \setminus \Delta$ is open, there exists a basic open set $U \times V$ in $X \times X$ such that:

Then $x \in U$ and $y \in V$. If $z \in U \cap V$, then $(z, z) \in U \times V \subseteq (X \times X) \setminus \Delta$, which contradicts $(z, z) \in \Delta$! Thus $U \cap V = \emptyset$. Hence $X$ is Hausdorff. $\blacksquare$

1. Regular Spaces

Moving up the hierarchy, the $T_3$ axiom separates points from closed sets.

Definition 5.4 (Regular Space & $T_3$ Space):

1. A topological space $X$ is called regular if for every closed set $F \subseteq X$ and every point $x \notin F$, there exist disjoint open sets $U, V \subseteq X$ such that:

1. A space is called a $T_3$ space if it is both regular and $T_1$.

Remark on Terminology: Some authors define $T_3$ to include $T_1$; under our standard modern convention, a regular space that is also $T_1$ is called $T_3$. Since singletons $\{y\}$ are closed in a $T_1$ space, separating $x$ from $\{y\}$ immediately gives disjoint neighborhoods, so:

2. Characterization via Closed Neighborhood Bases

Theorem 5.4 (Closed Neighborhood Base Characterization): A $T_1$ space $X$ is $T_3$ (regular) if and only if for every point $x \in X$ and every open neighborhood $U$ of $x$, there exists an open neighborhood $V$ of $x$ such that:

In other words, every point has a neighborhood base consisting of closed sets.

Proof: ($\Rightarrow$) Let $x \in U$ with $U$ open. The set $F = X \setminus U$ is closed, and $x \notin F$. By regularity, there exist disjoint open sets $V$ and $W$ such that $x \in V$ and $F \subseteq W$. Since $V \cap W = \emptyset$, we have $V \subseteq X \setminus W$. Since $X \setminus W$ is closed, $\operatorname{cl}(V) \subseteq X \setminus W$. Since $F \subseteq W$, we have $X \setminus W \subseteq X \setminus F = U$. Thus $x \in V \subseteq \operatorname{cl}(V) \subseteq U$.

($\Leftarrow$) Let $F$ be closed and $x \notin F$. Then $U = X \setminus F$ is an open neighborhood of $x$. By hypothesis, choose open $V$ with $x \in V \subseteq \operatorname{cl}(V) \subseteq U$. Let $W = X \setminus \operatorname{cl}(V)$. Then $W$ is open, $F = X \setminus U \subseteq X \setminus \operatorname{cl}(V) = W$, and $V \cap W = \emptyset$. Thus $x$ and $F$ are separated by disjoint open sets, so $X$ is regular. $\blacksquare$

1. Hereditary Properties

How do the axioms $T_0, T_1, T_2, T_3$ behave when taking subspaces?

Theorem 5.5 (Heredity of Lower Separation Axioms): Each of the separation properties $T_0, T_1, T_2$, and $T_3$ is hereditary: If $X$ satisfies $T_i$ ($i \in \{0, 1, 2, 3\}$), then every subspace $Y \subseteq X$ with the relative topology satisfies $T_i$.

Proof for $T_2$: Let $x \ne y \in Y$. Since $Y \subseteq X$, $x, y \in X$. Since $X$ is $T_2$, there exist disjoint open sets $U, V \subseteq X$ with $x \in U, y \in V$, and $U \cap V = \emptyset$. In the subspace topology on $Y$, $U_Y = U \cap Y$ and $V_Y = V \cap Y$ are open in $Y$. Moreover:

Thus $Y$ is Hausdorff. $\blacksquare$

2. Product Invariance

Theorem 5.6 (Product Invariance): Let $\{X_\alpha\}_{\alpha \in I}$ be an arbitrary non-empty family of topological spaces. The product space $X = \prod_{\alpha \in I} X_\alpha$ equipped with the product topology satisfies $T_i$ ($i \in \{0, 1, 2, 3\}$) if and only if every coordinate space $X_\alpha$ satisfies $T_i$.

Proof for $T_2$: ($\Leftarrow$) Let $x = (x_\alpha)$ and $y = (y_\alpha)$ be distinct points in $\prod X_\alpha$. Then there exists some coordinate $\beta \in I$ such that $x_\beta \ne y_\beta \in X_\beta$. Since $X_\beta$ is Hausdorff, there exist disjoint open sets $U_\beta, V_\beta \subseteq X_\beta$ separating $x_\beta$ and $y_\beta$. Consider the cylinders $U = \pi_\beta^{-1}(U_\beta)$ and $V = \pi_\beta^{-1}(V_\beta)$. Both $U$ and $V$ are open in the product topology, $x \in U$, $y \in V$, and:

Thus $\prod X_\alpha$ is Hausdorff. $\blacksquare$

1. The Line with Two Origins

A celebrated counterexample that shows local Euclidean structure does not guarantee the Hausdorff property!

Definition 5.5 (Line with Two Origins): Let $X = (\mathbb{R} \setminus \{0\}) \cup \{0_A, 0_B\}$ where $0_A \ne 0_B$ are two distinct points. Topologize $X$ by declaring:

• Any open interval in $\mathbb{R}$ not containing $0$ is open in $X$.
• Basic neighborhoods of $0_A$ are of the form $(-\varepsilon, 0) \cup \{0_A\} \cup (0, \varepsilon)$ for $\varepsilon > 0$.
• Basic neighborhoods of $0_B$ are of the form $(-\varepsilon, 0) \cup \{0_B\} \cup (0, \varepsilon)$ for $\varepsilon > 0$.

Theorem 5.7 (Properties of the Line with Two Origins):

1. $X$ is locally Euclidean (every point has a neighborhood homeomorphic to an open interval in $\mathbb{R}$).
2. $X$ is $T_1$ (singletons are closed).
3. $X$ is NOT Hausdorff ($T_2$)!

Proof that $X$ fails $T_2$: Consider the two origins $0_A \ne 0_B$. Let $U$ be any open neighborhood of $0_A$, and $V$ any open neighborhood of $0_B$. By definition, $U$ contains $(-\varepsilon_1, 0) \cup (0, \varepsilon_1)$ and $V$ contains $(-\varepsilon_2, 0) \cup (0, \varepsilon_2)$ for some $\varepsilon_1, \varepsilon_2 > 0$. Let $\delta = \min\{\varepsilon_1, \varepsilon_2\} > 0$. Then $(0, \delta) \subseteq U \cap V$. Thus $U \cap V \ne \emptyset$! The two origins $0_A$ and $0_B$ can never be separated by disjoint open sets. Hence $X$ is not Hausdorff. Consequently, the sequence $x_n = 1/n$ converges simultaneously to both $0_A$ and $0_B$! $\blacksquare$