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Separation Axioms

Separation axioms formalise how well points and closed sets can be “separated” by open sets.

Definition. XX is T0T_0 if for any two distinct points x,yXx, y \in X, at least one has an open neighbourhood not containing the other.

Example 8.1. The Sierpiński space {0,1}\{0, 1\} with topology {,{0},{0,1}}\{\emptyset, \{0\}, \{0, 1\}\} is T0T_0 but not T1T_1.

Definition. XX is T1T_1 if for any two distinct points x,yXx, y \in X, there exist open sets U,VU, V with xUx \in U, yUy \notin U and yVy \in V, xVx \notin V.

Proposition 8.1. XX is T1T_1 if and only if every singleton {x}\{x\} is closed.

Example 8.2. Any infinite set with the cofinite topology is T1T_1 but not T2T_2.

Definition. XX is Hausdorff (T2T_2) if for any two distinct points x,yXx, y \in X, there exist disjoint open sets U,VU, V with xUx \in U and yVy \in V.

Proposition 8.2. Every metric space is Hausdorff.

Proposition 8.3. In a Hausdorff space, every convergent sequence has a unique limit.

Proposition 8.4. T2    T1    T0T_2 \implies T_1 \implies T_0.

8.5 T3T_3 (Regular) and T4T_4 (Normal)

Section titled “8.5 T3T_3T3​ (Regular) and T4T_4T4​ (Normal)”

Definition. XX is regular (T3T_3) if it is T1T_1 and for any point xx and closed set FF with xFx \notin F, there exist disjoint open sets U,VU, V with xUx \in U and FVF \subseteq V.

Definition. XX is normal (T4T_4) if it is T1T_1 and for any two disjoint closed sets F1,F2F_1, F_2, there exist disjoint open sets U,VU, V with F1UF_1 \subseteq U and F2VF_2 \subseteq V.

Proposition 8.5. T4    T3    T2T_4 \implies T_3 \implies T_2 (assuming T1T_1).

Proposition 8.6. Every compact Hausdorff space is normal.

Theorem 8.1 (Urysohn’s Lemma). If XX is normal and F1,F2F_1, F_2 are disjoint closed subsets, then there exists a continuous function f:X[0,1]f : X \to [0, 1] with fF1=0f|_{F_1} = 0 and fF2=1f|_{F_2} = 1.

This is a fundamental tool in topology, used to construct partitions of unity and to prove extension theorems.

Definition. A space XX is completely regular (Tychonoff, T3.5T_{3.5}) if it is T1T_1 and for every point xx and closed set FF with xFx \notin F, there exists a continuous function f:X[0,1]f : X \to [0, 1] with f(x)=0f(x) = 0 and f(F)=1f(F) = 1.

Proposition 8.7. Every normal space is completely regular. Every completely regular space is regular: T4    T3.5    T3T_4 \implies T_{3.5} \implies T_3.

Example 8.3. The Sorgenfrey line (lower limit topology on R\mathbb{R}) is Tychonoff but not normal.

Theorem 8.2 (Tietze Extension Theorem). If XX is normal, FXF \subseteq X is closed, and f:F[a,b]f : F \to [a, b] is continuous, then ff extends to a continuous function f~:X[a,b]\tilde{f} : X \to [a, b] with f~F=f\tilde{f}|_F = f.

This follows from Urysohn’s lemma by constructing a sequence of approximations whose sum converges uniformly to the extension.

Theorem 8.3 (Urysohn Metrization Theorem). Every second-countable regular T1T_1 space is metrizable (its topology is induced by some metric).

Proof sketch. Embed XX into the Hilbert cube [0,1]ω[0, 1]^\omega using a countable collection of Urysohn functions constructed from a countable basis. The Hilbert cube is metrizable, so the subspace topology on XX is metrizable. \blacksquare

Corollary 8.8. A space is a separable metric space if and only if it is second-countable and regular T1T_1.

The hierarchy of separation axioms:

T4    T3.5    T3    T2    T1    T0T_4 \implies T_{3.5} \implies T_3 \implies T_2 \implies T_1 \implies T_0

Each implication is strict: there exist spaces satisfying each level but not the next.

Problem 1. Show that the Zariski topology on Rn\mathbb{R}^n is T1T_1 but not T2T_2.

Problem 2. Prove that a closed subspace of a normal space is normal.

Solution. Let XX be normal and FXF \subseteq X closed. Let A,BA, B be disjoint closed sets in FF. Then A,BA, B are also closed in XX (since FF is closed). By normality of XX, there exist disjoint open U,VXU, V \subseteq X with AUA \subseteq U, BVB \subseteq V. Then UFU \cap F and VFV \cap F are disjoint open in FF separating AA and BB. \blacksquare

Problem 3. Show that every metric space is normal.

Problem 4. Prove that a product of two T2T_2 spaces is T2T_2. Is the same true for T3T_3 or T4T_4?

Problem 5. Show that the Sorgenfrey plane R×R\mathbb{R}_\ell \times \mathbb{R}_\ell is not normal, even though the Sorgenfrey line R\mathbb{R}_\ell is normal. (Hint: the antidiagonal {(x,x):xR}\{(x, -x) : x \in \mathbb{R}\} is closed and discrete.)

8.12 T5T_5 (Completely Normal) and T6T_6 (Perfectly Normal)

Section titled “8.12 T5T_5T5​ (Completely Normal) and T6T_6T6​ (Perfectly Normal)”

Definition. A space XX is completely normal (T5T_5) if every subspace of XX is normal. Equivalently, for any two separated sets A,BA, B (i.e., AˉB=ABˉ=\bar{A} \cap B = A \cap \bar{B} = \emptyset), there exist disjoint open sets separating them.

Definition. A space XX is perfectly normal (T6T_6) if XX is normal and every closed set is a GδG_\delta set (a countable intersection of open sets).

Proposition 8.9. Every metric space is T6T_6 (perfectly normal).

Proposition 8.10. T6    T5    T4T_6 \implies T_5 \implies T_4 (assuming T1T_1).

Theorem 8.4 (Tychonoff Product Theorem). The product of any collection of compact topological spaces is compact (in the product topology).

This is one of the most important theorems in general topology, equivalent to the axiom of choice. The proof uses the finite intersection property and Zorn’s lemma.

For a Tychonoff space XX, the Stone-Cech compactification βX\beta X is the unique compact Hausdorff space containing XX as a dense subspace such that every continuous map from XX to a compact Hausdorff space extends continuously to βX\beta X.

Example. βN\beta \mathbb{N} is the set of ultrafilters on N\mathbb{N} with the topology generated by {A:AN}\{A^* : A \subseteq \mathbb{N}\} where A={UβN:AU}A^* = \{U \in \beta\mathbb{N} : A \in U\}.

Problem 6. Prove that every compact subset of a Hausdorff space is closed.

Problem 7. Show that Q\mathbb{Q} with the subspace topology from R\mathbb{R} is not a T4T_4 space.

Problem 8. Prove that a space is Hausdorff if and only if the diagonal Δ={(x,x):xX}\Delta = \{(x, x) : x \in X\} is closed in X×XX \times X.