Hausdorff Implies T1

Statement

A topological space $X$ is Hausdorff (or $T_2$) if for every two distinct points $x \ne y$ there exist disjoint open sets $U \ni x$ and $V \ni y$:

A space is T1 if every singleton $\{x\}$ is a closed set:

Theorem: Every Hausdorff space is T1:

The converse fails: the cofinite topology on an infinite set is T1 (singletons are closed since their complements are cofinite, hence open) but not Hausdorff (any two nonempty open sets intersect in an infinite set).

Visualization

Hausdorff separation: two points $x$ and $y$ separated by disjoint open neighbourhoods.

  X:
     ┌──────────────────────────────┐
     │   ┌─────┐         ┌─────┐   │
     │   │  U  │         │  V  │   │
     │   │  x  │         │  y  │   │
     │   └─────┘         └─────┘   │
     │       U ∩ V = ∅              │
     └──────────────────────────────┘

  T1 conclusion: {y} is closed.
  Proof: X \ {y} = ⋃_{x ≠ y} U_x
         where U_x is open and misses y.
         This union is open, so {y} is closed.

Table: separation axioms

Axiom	Condition	Implication
$T_0$ (Kolmogorov)	for $x \ne y$, some open distinguishes them	weakest
$T_1$ (Fréchet)	singletons are closed	$T_2 \Rightarrow T_1 \Rightarrow T_0$
$T_2$ (Hausdorff)	distinct points have disjoint open nbhds	intermediate
$T_3$ (Regular)	point and closed set have disjoint open nbhds	$T_3 \Rightarrow T_2$
$T_4$ (Normal)	disjoint closed sets have disjoint open nbhds	strongest common

Proof Sketch

1. Let $X$ be Hausdorff and fix $y \in X$. We show $\{y\}$ is closed by showing $X \setminus \{y\}$ is open.

2. Take any $x \in X \setminus \{y\}$, so $x \ne y$. By the Hausdorff condition, there exist open sets $U_x \ni x$ and $V_x \ni y$ with $U_x \cap V_x = \emptyset$.

3. In particular, $y \notin U_x$ (since $U_x \cap V_x = \emptyset$ and $y \in V_x$), so $U_x \subseteq X \setminus \{y\}$.

4. Then $X \setminus \{y\} = \bigcup_{x \ne y} U_x$ is a union of open sets, hence open.

5. Therefore $\{y\}$ is closed. Since $y$ was arbitrary, $X$ is T1.

Connections

• Urysohn's LemmaUrysohn's Lemma$$\exists\, f : X \to [0,1],\quad f|_s = 0,\; f|_t = 1$$In a normal space, disjoint closed sets can be separated by a continuous functionRead more → — Urysohn's lemma requires normality (T4), which implies Hausdorff (T2), which implies T1; the lemma produces continuous functions separating closed sets, generalising the separation axiom hierarchy.
• Heine–Borel TheoremHeine–Borel Theorem$$K \subseteq \mathbb{R}^n \;\text{compact} \iff K \;\text{closed and bounded}$$In ℝⁿ, a subset is compact if and only if it is closed and boundedRead more → — $\mathbb{R}^n$ is Hausdorff (metric spaces are always Hausdorff), so the Heine–Borel characterisation of compact sets as closed and bounded takes place in a T1 space where singletons are automatically closed.
• Compact Subset of Hausdorff is ClosedCompact Subset of Hausdorff is Closed$$X \text{ Hausdorff},\; K \text{ compact} \Rightarrow K \text{ closed}$$In a Hausdorff space every compact subset is closed, generalising the Heine–Borel direction compact implies closedRead more → — the T1 property (singletons closed) is necessary but not sufficient for compact sets to be closed; Hausdorff (T2) is the right axiom, as proven in the companion note.
• Bolzano–Weierstrass TheoremBolzano–Weierstrass Theorem$$\text{bounded } (x_n) \subseteq \mathbb{R}^n \implies \exists\, \text{convergent subsequence}$$Every bounded sequence in ℝⁿ has a convergent subsequenceRead more → — convergent sequences in Hausdorff spaces have unique limits (the Hausdorff condition separates any two distinct limit candidates), a key feature that T1 alone does not guarantee.

import Mathlib.Topology.Separation.Hausdorff

/-- Every Hausdorff space is T1: singletons are closed.
    Mathlib provides this as an instance with priority 100. -/
theorem hausdorff_implies_t1
    (X : Type*) [TopologicalSpace X] [T2Space X] : T1Space X :=
  inferInstance