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}.