Mathematics.

point set topology

Separation Axioms

Topology40 minDifficulty7 out of 10

You should know: hausdorff spaces

Overview

The separation axioms are a hierarchy of conditions (T0, T1, T2, T3, T4, ...) that a topological space may or may not satisfy, each demanding progressively stronger ways of 'telling points or sets apart' using open sets. T0 (Kolmogorov) merely asks that for any two distinct points, some open set contains one but not the other; T1 strengthens this to say singletons are closed; T2 (Hausdorff) demands disjoint open neighborhoods for distinct points; T3 (regular, plus T1) separates points from closed sets not containing them; and T4 (normal, plus T1) separates disjoint closed sets from each other. Each axiom in the hierarchy strictly implies the ones before it (T4+T1 ⟹ T3+T1 ⟹ T2 ⟹ T1 ⟹ T0), and standard counterexamples show none of the reverse implications hold — the hierarchy genuinely stratifies how 'nicely separated' a space's points and closed sets are.

Intuition

Think of the separation axioms as a ladder of increasingly demanding etiquette rules for how open sets must treat points and closed sets. T0 is the weakest possible manners: given two different points, SOME open set must at least distinguish them (contain one but not the other) — otherwise the points would be topologically identical twins, indistinguishable by any open set. T1 insists that individual points can always be isolated as closed sets (every finite set is closed). T2 (Hausdorff) is the familiar 'two points, two disjoint neighborhoods' rule. T3 (regular) extends the courtesy to closed sets: not just two points, but a point and a closed set not containing it can be separated by disjoint open sets — imagine a person and a wall they're not touching, each getting their own bubble of space. T4 (normal) goes further still: even two disjoint closed sets (not just a point and a closed set) can be wrapped in disjoint open neighborhoods, like two separate groups of people in a crowd who never have to touch, no matter how you draw the boundary around each group.

Formal Definition

Definition

Let (X, τ) be a topological space. The separation axioms are, in increasing strength:

T0: ∀x≠y, ∃U∈τ containing exactly one of x,yT_0: \ \forall x\neq y,\ \exists U \in \tau \text{ containing exactly one of } x, y

Some open set distinguishes any two distinct points, but need not contain BOTH separately

T0 (Kolmogorov)
T1: ∀x≠y, ∃U∋x open with y∉U(equivalently, every singleton {x} is closed)T_1: \ \forall x \neq y,\ \exists U \ni x \text{ open with } y \notin U \quad(\text{equivalently, every singleton } \{x\} \text{ is closed})

Each point can be excluded from a neighborhood of the other point, symmetrically

T1 (Fréchet)
T2: ∀x≠y, ∃U,V∈τ disjoint with x∈U, y∈VT_2: \ \forall x\neq y,\ \exists U,V \in \tau \text{ disjoint with } x\in U,\ y \in V

Distinct points have disjoint open neighborhoods

T2 (Hausdorff)
T3+T1=regular: ∀x∉C (C closed), ∃U,V∈τ disjoint with x∈U, C⊆VT_3 + T_1 = \text{regular}: \ \forall x \notin C \text{ (C closed)},\ \exists U,V \in \tau \text{ disjoint with } x \in U,\ C \subseteq V

A point outside a closed set can be separated from it by disjoint open sets

Regular (T3, combined with T1)
T4+T1=normal: ∀ disjoint closed C,D, ∃U,V∈τ disjoint with C⊆U, D⊆VT_4 + T_1 = \text{normal}: \ \forall \text{ disjoint closed } C, D,\ \exists U, V \in \tau \text{ disjoint with } C\subseteq U,\ D\subseteq V

Any two disjoint closed sets can be separated by disjoint open sets

Normal (T4, combined with T1)

Notation

NotationMeaning
Ti, i=0,1,2,3,4T_i,\ i=0,1,2,3,4Standard labels for the separation axioms, increasing in strength with i (roughly — T3 and T4 are conventionally combined with T1 to get 'regular' and 'normal')
regular=T3+T1\text{regular} = T_3 + T_1A space satisfying both the T3 axiom and T1 (so points are closed AND separable from closed sets)
normal=T4+T1\text{normal} = T_4 + T_1A space satisfying both T4 and T1 (disjoint closed sets separable, and points are closed)

Derivation

Deriving that normal + T1 (i.e. T4+T1) implies regular + T1 (T3+T1), by treating a single point as a special case of a closed set.

Suppose X is normal (T4) and T1.\text{Suppose } X \text{ is normal (T4) and } T_1.

Assume the hypotheses

Let x∈X, C⊆X closed, x∉C.\text{Let } x \in X, \ C \subseteq X \text{ closed}, \ x \notin C.

Set up the regularity scenario: a point outside a closed set

Since X is T1,{x} is closed.\text{Since } X \text{ is } T_1, \{x\} \text{ is closed.}

T1 exactly guarantees singletons are closed

{x} and C are disjoint closed sets (since x∉C), so normality (T4) gives disjoint open U⊇{x},V⊇C.\{x\} \text{ and } C \text{ are disjoint closed sets (since } x \notin C\text{), so normality (T4) gives disjoint open } U \supseteq \{x\}, V\supseteq C.

Apply the T4 (normal) separation directly to the two disjoint closed sets {x} and C

∴x∈U, C⊆V, U∩V=∅ — exactly the T3 (regularity) condition.\therefore x \in U,\ C \subseteq V,\ U\cap V=\emptyset \text{ — exactly the T3 (regularity) condition.}

This is precisely the regularity axiom, so normal+T1 implies regular+T1

Proofs

Every Hausdorff (T2) space is T1
  1. Let X be Hausdorff and x≠y∈X.\text{Let } X \text{ be Hausdorff and } x \neq y \in X.(Assume the T2 hypothesis and pick arbitrary distinct points)
  2. By T2, ∃U,V open, disjoint, with x∈U, y∈V.\text{By T2, } \exists U, V \text{ open, disjoint, with } x\in U,\ y \in V.(Definition of Hausdorff)
  3. In particular, U is open with x∈U and y∉U (since U∩V=∅ and y∈V).\text{In particular, } U \text{ is open with } x \in U \text{ and } y \notin U \text{ (since } U\cap V=\emptyset \text{ and } y \in V\text{).}(Disjointness of U and V means y, being in V, cannot also be in U)
  4. This is exactly the T1 condition (with roles of x,y symmetric via V).\text{This is exactly the T1 condition (with roles of } x, y \text{ symmetric via V).}(T1 requires: for each ordered pair (x,y) of distinct points, an open set containing x but not y — which U supplies)
  5. ∴X is T1.\therefore X \text{ is } T_1.(Both directions (x excluded from y's neighborhood and vice versa) are witnessed by U and V respectively)

Properties

Hierarchy of implications

normal  ⟹  regular  ⟹  T2  ⟹  T1  ⟹  T0\text{normal} \implies \text{regular} \implies T_2 \implies T_1 \implies T_0

Condition: Each stronger axiom in the chain implies all weaker ones (given the T1 pairing for regular/normal); none of the reverse implications hold in general.

Metric spaces are normal

Every metric space is T4 (normal) and T1, hence also regular and Hausdorff\text{Every metric space is } T_4 \text{ (normal) and } T_1, \text{ hence also regular and Hausdorff}

Example: Disjoint closed sets C, D in a metric space can be separated using the function d(x,C) − d(x,D), whose sign gives disjoint open preimages.

T1 iff singletons are closed

X is T1  ⟺  {x} is closed for every x∈XX \text{ is } T_1 \iff \{x\} \text{ is closed for every } x \in X

Condition: Direct consequence of unwinding the T1 definition symmetrically for both x and y.

Finite T1 spaces are discrete

X finite and T1  ⟹  X has the discrete topologyX \text{ finite and } T_1 \implies X \text{ has the discrete topology}

Condition: Every singleton is closed, so every subset (a finite union of singletons) is closed, hence also open by taking complements.

Cofinite topology is T1 but not T2

On an infinite set X, the cofinite topology is T1 but not Hausdorff\text{On an infinite set } X, \text{ the cofinite topology is } T_1 \text{ but not Hausdorff}

Example: Every singleton has finite (hence closed) complement, so singletons are closed (T1); but any two nonempty open sets in the cofinite topology have finite complements, so they must intersect (their union would need infinite complement, impossible for two cofinite sets to avoid overlapping) — no disjoint open neighborhoods exist, failing T2.

Applications

Domain theory and denotational semantics use weaker separation axioms (often only T0) deliberately, since many useful order-theoretic topologies (like the Scott topology) are not even Hausdorff, yet still support a rich theory of computation and approximation.

Worked Examples

  1. T0 requires some open set containing exactly one of a, b. The only open sets are ∅ and X={a,b}, neither of which contains exactly one of a,b.

    τ={∅,{a,b}}\tau = \{\emptyset, \{a,b\}\}

  2. No open set separates a from b in this way, so the space fails T0.

    ∄ U∈τ with ∣U∩{a,b}∣=1\nexists\, U \in \tau \text{ with } |U \cap \{a,b\}| = 1

Answer: No — the indiscrete topology on a set with 2+ points fails even the weakest separation axiom, T0.

Practice Problems

Difficulty 5/10

Which separation axiom is equivalent to saying every singleton {x} is closed?

Difficulty 6/10

Give an example of a topological space that is T1 but not T2 (Hausdorff).

Difficulty 8/10

Prove that every finite T1 space is discrete (every subset is open).

Quiz

The T4 (normal) separation axiom separates:
Which chain of implications correctly describes the separation axiom hierarchy (with T1 combined where needed)?

Summary

  • The separation axioms T0, T1, T2 (Hausdorff), T3+T1 (regular), T4+T1 (normal) form a strict hierarchy of increasingly strong 'point/set separation' conditions on a topological space.
  • T0: some open set distinguishes any two points. T1: singletons are closed. T2: distinct points have disjoint open neighborhoods.
  • Regular (T3+T1) separates points from closed sets not containing them; normal (T4+T1) separates disjoint closed sets from each other.
  • Every metric space is normal (hence regular, Hausdorff, T1, T0), via the distance-difference function construction.
  • The cofinite topology on an infinite set is a standard example that is T1 but not T2, showing the hierarchy is genuinely strict.

References

  1. WebsiteWikipedia — Separation axiom
  2. WebsiteWikipedia — Normal space
  3. BookMunkres, J. Topology, 2nd ed. Ch. 4.