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Every completely regular space is regular, and every Tychonoff space is $T_{3}$

Statement

Let $(X, T)$ be a topological space (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison). If $X$ is completely regular (Completely regular spaces and Tychonoff ( $T_{3 \frac{1}{2}}$ ) spaces) then $X$ is regular (Regular spaces and $T_{3}$ spaces, with the source disagreement over whether regularity includes $T_{1}$ stated explicitly). Consequently every Tychonoff space is $T_{3}$ , being completely regular and $T_{1}$ ( $T_{0}$ (Kolmogorov) and $T_{1}$ (Frechet) spaces).

This page does not prove the converse and does not assert it: a regular space that is not completely regular would need a construction this page does not carry, so whether the implication reverses is left open here.

Facts & Assumptions

Given: A completely regular space $(X, T)$ , a closed set $C \subseteq X$ and a point $x_{0} \in X ∖ C$ .

[A1]

Complete regularity supplies a continuous $f : X \to [0, 1]$ with $f (x_{0}) = 1$ and $f (y) = 0$ for every $y \in C$ (Completely regular spaces and Tychonoff ( $T_{3 \frac{1}{2}}$ ) spaces).

[A2]

$X$ is regular when every such pair $(C, x_{0})$ admits disjoint open $U ∋ x_{0}$ and $V \supseteq C$ (Regular spaces and $T_{3}$ spaces, with the source disagreement over whether regularity includes $T_{1}$ stated explicitly).

[L1]

A map into the subspace $[0, 1]$ of $R$ is continuous exactly when it is continuous as a map into $R$ , and the open subsets of $[0, 1]$ are the traces on $[0, 1]$ of the open subsets of $R$ (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

[L2]

$f^{- 1} [W]$ is open in $X$ for every open $W$ in the target (For a map of spaces the following agree: continuity at every point, preimages of open sets open, preimages of closed sets closed, preimages of subbasic open sets open, and $f (\overline{A}) \subseteq \overline{f (A)}$ , clause (b), Continuity of a map of topological spaces at a point and globally).

[L3]

The sets $(1 / 2, \infty)$ and $(- \infty, 1 / 2)$ are open in the usual topology of $R$ , they are disjoint, and $0 < 1 / 2 < 1$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, The absolute value makes $R$ a metric space: $d (x, y) = ∣ x - y ∣$ is a metric, its open balls are the intervals $(x - r, x + r)$ , and it is unbounded, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not).

Proof

technique · direct

1.1A1L1L3

Fix $f$ as in [A1], and put $W_{1} : = (1 / 2, \infty) \cap [0, 1]$ and $W_{0} : = (- \infty, 1 / 2) \cap [0, 1]$ , which are open in $[0, 1]$ and disjoint by [L1] and [L3].

1.2A1L2

Put $U : = f^{- 1} [W_{1}]$ and $V : = f^{- 1} [W_{0}]$ ; both are open in $X$ by [L2].

2.1step 1.1step 1.2A1L3

$x_{0} \in U$ , since $f (x_{0}) = 1 > 1 / 2$ and $1 \in [0, 1]$ .

2.2step 1.1step 1.2A1L3

$C \subseteq V$ , since $f (y) = 0 < 1 / 2$ and $0 \in [0, 1]$ for every $y \in C$ .

2.3step 1.1step 1.2L3

$U \cap V = \emptyset$ : a point of both would satisfy $f (x) > 1 / 2$ and $f (x) < 1 / 2$ , which is impossible by trichotomy of the order of $R$ .

3.1step 1.2step 2.1step 2.2step 2.3A2

By steps 1.2, 2.1, 2.2 and 2.3 the pair $(C, x_{0})$ is separated by disjoint open sets, and since $C$ and $x_{0}$ were arbitrary, $X$ is regular by [A2].

4.1step 3.1A2∎

If in addition $X$ is $T_{1}$ then $X$ is regular and $T_{1}$ , that is $T_{3}$ ; so every Tychonoff space is $T_{3}$ .

Remarks

The threshold $1 / 2$ is arbitrary. Any $t$ with $0 < t < 1$ works, and the same two-set construction applied to a function with values in $R$ rather than $[0, 1]$ gives the same conclusion; the normalisation of Completely regular spaces and Tychonoff ( $T_{3 \frac{1}{2}}$ ) spaces is used only to know that $f (x_{0})$ and the values on $C$ lie on opposite sides of the threshold.
What the theorem does not give. It says nothing about separating two closed sets, and complete regularity does not imply normality. In the other direction, a normal $T_{1}$ space is completely regular, but that is Urysohn's lemma and is the one arrow of the classical chain this page cannot reach (Conventions on this page, and the one implication of the classical chain that is not available at this point in the reading order).

Every completely regular space is regular, and every Tychonoff space is $T_{3}$

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

Every completely regular space is regular, and every Tychonoff space is T3T3​

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

Every completely regular space is regular, and every Tychonoff space is $T_{3}$