# General Topology – Elementary Separation Axioms

This blog post discusses the elementary separation axioms ${T_0}$, ${T_1}$, and ${T_2}$. There are also the more sophisticated separation axioms ${T_3}$, ${T_{3{1\over2}}}$, and ${T_4}$, which are explained (but not numbered) in Section 4-2 of Munkres’s Topology.

The separation axioms ${T_0}$, ${T_1}$, and ${T_2}$ are properties that a specific topological space ${X}$ may or may not have. By giving examples, we will see that the condition that a space be ${T_{i+1}}$ is strictly stronger than the condition that a space be ${T_i}$. Finally, we will discuss how the properties ${T_i}$ are inhertied by the constructions on topological spaces that we know, namely taking subspaces and product spaces.

The Axioms. Here are the axioms. We say a space ${X}$ is ${T_i}$ when:

• ${T_0}$: For any two distinct points ${x, y \in X}$, either there is an open set around ${y}$ not containing ${x}$ (or both).
• ${T_1}$: For any two distinct points ${x, y \in X}$, there is an open set around ${x}$ not containing ${y}$ and there is an open set around ${y}$ not containing ${x}$.
• ${T_2}$: For any two distinct points ${x, y \in X}$, there are disjoint open sets ${U_x}$ and ${U_y}$ with ${x \in U_x}$ and ${y \in U_y}$.

The axiom ${T_2}$ is especially important and is also called the Hausdorff axiom. We say a space is Hausdorf if it is satisfies ${T_2}$. We know some examples of Hausdorff spaces. For example a discrete space is Hausdorff. More generally, metric spaces are Hausdorff.

Theorem. If ${(X, d)}$ is a metric space, then ${X}$ is Hausdorff.

Proof. Let ${x, y \in X}$ be distinct points. Let ${r = d(x, y)/2}$. The balls ${B_r(x)}$ and ${B_r(y)}$ are disjoint (by the triangle inequality) open sets containing ${x}$ and ${y}$ respectively.

Clearly ${T_2 \implies T_1 \implies T_0}$. Also, clearly not all spaces satisfy even the weakest axiom ${T_0}$: an indiscrete space with more than one point does not satisfy ${T_0}$. Let us give some examples to show that ${T_0}$ does not imply ${T_1}$, and ${T_1}$ does not imply ${T_2}$.

Example 1. A space that is ${T_0}$ but not ${T_1}$. Consider the space whose open sets are given by the “dot-diagram” below.

There is an open set around the point on the left not containing the point on the right, but not vice-versa.

Example 2. A space that is ${T_1}$ but not ${T_2}$. Let ${X}$ be any infinite set with the cofinite topology. Thus, a subset ${U}$ is open in ${X}$ if and only if either ${U}$ is empty or ${X \setminus U}$ is finite. For distinct points ${x, y \in X}$, ${X \setminus\{y\}}$ and ${X \setminus \{x\}}$ are open sets containing ${x}$ but not contaning ${y}$, and containing ${y}$ but not containing ${x}$ respectively, so ${X}$ is ${T_1}$. On the other hand, let ${U_x}$ and ${U_y}$ be any open sets containing ${x}$ but not containing ${y}$, and containing ${y}$ but not containing ${x}$ respectively. Then ${X \setminus U_x}$ and ${X \setminus U_y}$ are finite sets and so

$\displaystyle X \setminus (U_x \cap U_y) = (X \setminus U_x) \cap (X \setminus U_y)$

is a finite set. Since ${X}$ is infinite, we must have that ${U_x \cap U_y}$ is not empty. Since ${U_x}$ and ${U_y}$ were arbitrary, ${X}$ is not ${T_2}$.

Example 3. Another example of a space that is ${T_1}$ but not ${T_2}$ is ${\mathbb{C}^n}$ with the Zariski topology. Recall that the Zariski topology has as a basis the set

$\displaystyle \{U_f: f \text{ is a polynomial in }z_1, \dots, z_n\}$

where

$\displaystyle U_f = \{(x_1, \dots, x_n) \in \mathbb{C}^n : f(x_1, \dots, x_n) \neq 0\}.$

Given ${{\bf x} = (x_1, \dots, x_n)}$ and ${{\bf y} = (y_1, \dots, y_n)}$, distinct points in ${\mathbb{C}^n}$, the open sets

$\displaystyle U = U_{(z_1 - y_1)} \cup \dots \cup U_{(z_n - y_n)} = \mathbb{C}^n \setminus \{{\bf y}\}$

$\displaystyle V = U_{(z_1 - x_1)} \cup \dots \cup U_{(z_n - x_n)} = \mathbb{C}^n \setminus \{{\bf x}\}$

contain ${{\bf x}}$ but not ${{\bf y}}$ and ${{\bf y}}$ but not ${{\bf x}}$ respectively. This shows that this topology is ${T_1}$. To see that the topology is not ${T_2}$, it suffices to show that the intersection of nonempty open sets is nonempty. For arbitrary open sets ${U}$ and ${V}$, write

$\displaystyle U = \bigcup U_{f_\alpha},\text{ }V = \bigcup U_{g_\beta}.$

Then ${U \cap V = \bigcup\left(U_{f_\alpha} \cap U_{g_\beta}\right)}$, so it suffices to show that the intersection of two nonempty basic open sets is nonempty, i.e., that given nonzero polynomials ${f}$ and ${g}$, there is a point ${{\bf x}}$ such that ${f({\bf x})}$ and ${g({\bf x})}$ are both nonzero. This is clear if ${n = 1}$ (since ${f}$ and ${g}$ each ahve ony finitely many zeros) and can be shown for general ${n}$ by induction: for all but at most finitely many values of ${\lambda}$, ${f(z_1, \dots, z_{n-1}, \lambda)}$ and ${g(z_1, \dots, z_{n-1}, \lambda)}$ are nonzero polynomials in ${z_1, \dots, z_{n-1}}$.

Note that both examples of ${T_1}$ spaces that are not ${T_2}$ are infinite. It turns out that a finite ${T_1}$ space must be discrete. Let ${X}$ be a ${T_1}$ topological space whose underlying set is a finite set. For any ${x \in X}$ the ${T_1}$ property implies that the intersection of all open sets containing ${x}$ is just the set ${\{x\}}$. There are only a finite number of open sets containing ${x}$ (since there are only finitely many subsets of ${X}$), so eveyr one point set is open; this implies that ${X}$ is discrete. This observation is closely related to the following fact.

Proposition. A topological space ${X}$ is ${T_1}$ if and only if every one point subset is closed.

Proof. If ${x}$ is ${T_1}$ and ${x \in X}$, then the ${T_1}$ property implies that for each ${y \in X}$ not equal to ${x}$, we can find an open set ${U_y}$ containing ${y}$ and not containing ${x}$. Then ${X \setminus \{x\} = \bigcup U_y}$ is an open set, and so ${\{x\}}$ is closed. Suppose on the other hand that eveyr one point set of ${X}$ is closed. Then for distinct points ${x, y \in X}$, ${U_x = X \setminus \{y\}}$ is an open set containing ${x}$ but containing ${y}$ and ${U_y = X \setminus\{x\}}$ is an open set containing ${y}$ but not containing ${x}$. Hence ${X}$ is ${T_1}$.

Finally, let us talk about subspaces and product spaces.

Theorem. Let ${X}$ be a topological space, and let ${A}$ be a subspace. If ${X}$ is ${T_0}$, ${T_1}$, or ${T_2}$ then so is ${A}$.

Proof. Let ${x}$ and ${y}$ be distinct points in ${A}$. Then ${x}$ and ${y}$ are also distinct points in ${X}$, and if ${X}$ is ${T_0}$, we can find an open subset ${U}$ of ${X}$ around one not containing the other. Then ${U \cup A}$ is an open subset of ${A}$ containing (the same) one of ${x}$ or ${y}$ and not the other, and so ${A}$ is ${T_0}$. If ${X}$ is ${T_1}$, we can find open subset ${U_x}$ and ${U_y}$ of ${X}$ such that ${x \in U_x}$, ${x \notin U_y}$, ${y \in U_y}$, and ${y \notin U_x}$. Then ${V_x = U_x \cap A}$ and ${V_y = U_y \cap A}$ are open subset of ${A}$ containing ${x}$ but not ${y}$ and containing ${y}$ but not ${x}$, respectively, and so ${A}$ is ${T_1}$. If ${X}$ is ${T_2}$, then we can choose ${U_x}$ and ${U_y}$ to be disjoint. Then ${V_x}$ and ${V_y}$ are disjoint, and so ${A}$ is ${T_2}$.

Theorem. Let ${X}$ and ${Y}$ be topological spaces. If ${X}$ and ${Y}$ are both ${T_0}$, ${T_1}$, or ${T_2}$ then so is ${X \times Y}$.

Proof. Let ${(x_1, y_1)}$ and ${(x_2, y_2)}$ be disjoint points in ${X \times Y}$. Then either ${x_1 \neq x_2}$ or ${y_1 \neq y_2}$. We will treat the case ${x_1 \neq x_2}$; the other case is entirely similar. If ${X}$ is ${T_0}$, then we can find an open subset ${U}$ of ${X}$ containing one of ${x_1}$ or ${x_2}$ but not the other. Then ${U \times Y}$ is an open subset of ${X \times Y}$ containing one of ${(x_1, y_1)}$ or ${(x_2, y_2)}$ but not the other, and so ${X \times Y}$ is ${T_0}$. If ${X}$ is ${T_1}$, we can find open subset ${U_1}$ and ${U_2}$ of ${X}$ such that ${x_1 \in U_1}$, ${x_1 \notin U_2}$, ${x_2 \in U_2}$, and ${y \notin U_1}$. Then ${U_1 \times Y}$ and ${U_2 \times Y}$ are open subset of ${X \times Y}$ containing ${(x_1, y_1)}$ but not ${(x_2, y_2)}$ and containing ${(x_2, y_2)}$ but not ${(x_1, y_1)}$, respectively, and so ${X \times Y}$ is ${T_1}$. If ${X}$ is ${T_2}$, then we can choose ${U_1}$ and ${U_2}$ to be disjoint. Then ${U_1 \times Y}$ and ${U_2 \times Y}$ are disjoint, and so ${X \times Y}$ is ${T_2}$.