TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29

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In a metric space every closed set is a zero set and a $G_{δ}$ , and the distance function separates a point from a closed set, so every metrizable space is Tychonoff and perfectly normal

Statement

Let $(X, d)$ be a metric space (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric) with its metric topology (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement), and write $1 / (n + 1)$ for the inverse of the canonical natural $ι (n + 1)$ of $R$ (The canonical natural $ι (n) = n \cdot 1_{F}$ of a field). Then:

Every closed set is a zero set. For closed $C \subseteq X$ there is a continuous $f : X \to R$ with $C = Z (f)$ (Zero sets and cozero sets of continuous real-valued functions); for $C \neq \emptyset$ one may take $f (x) = d (x, C)$ (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space), and for $C = \emptyset$ the constant function $1$ .
Every closed set is a $G_{δ}$ ( $G_{δ}$ and $F_{σ}$ subsets of a topological space, agreeing with the real-line notion): for $C \neq \emptyset$ , $C = ⋂_{n \in N} {x \in X : d (x, C) < 1 / (n + 1)},$ an intersection of open sets, and $\emptyset$ is open hence a $G_{δ}$ .
$X$ is completely regular (Completely regular spaces and Tychonoff ( $T_{3 \frac{1}{2}}$ ) spaces): for closed $C$ and $x_{0} \notin C$ the function $f (x) : = \min {1, d (x, C) / r}$ with $r : = d (x_{0}, C)$ is continuous, takes the value $1$ at $x_{0}$ and the value $0$ on $C$ , when $C \neq \emptyset$ ; for $C = \emptyset$ the constant function $1$ serves.
Consequently every metrizable space (Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not) is Tychonoff and perfectly normal, and hence $T_{6}$ , $T_{5}$ , $T_{4}$ , $T_{3 \frac{1}{2}}$ , $T_{3}$ , $T_{2 \frac{1}{2}}$ , $T_{2}$ , $T_{1}$ and $T_{0}$ .

No choice principle is used anywhere below.

Facts & Assumptions

Given: A metric space $(X, d)$ , a closed set $C \subseteq X$ , a point $x_{0} \in X ∖ C$ , and $R$ with its usual topology.

[L1]

For nonempty $S \subseteq X$ the distance $d (x, S)$ is defined, is $\geq 0$ , and $\overline{S} = {x : d (x, S) = 0}$ (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space, The closure of a nonempty $A$ is ${x : d (x, A) = 0}$ , equals $A$ together with its limit points, and is the smallest closed superset, claim 1).

[L2]

$∣ d (x, S) - d (y, S) ∣ \leq d (x, y)$ for nonempty $S$ ( $∣ d (x, A) - d (y, A) ∣ \leq d (x, y)$ , so the distance to a fixed nonempty set is $1$ -Lipschitz).

[L3]

A map between metric spaces satisfying an inequality $∣ g (x) - g (y) ∣ \leq L d (x, y)$ with $L > 0$ is continuous in the $ε$ - $δ$ sense, by $δ : = ε / L$ , and is therefore continuous as a map of topological spaces (Continuity of a map between metric spaces, at a point and globally, in the $ε$ - $δ$ form, Metric continuity characterisations, with countable choice for the sequential converse, clause (b), Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not).

[L4]

A set is closed exactly when it equals its closure (The closure of a nonempty $A$ is ${x : d (x, A) = 0}$ , equals $A$ together with its limit points, and is the smallest closed superset, claim 3); $\emptyset$ and $X$ are open (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

[L5]

For every real $ε > 0$ there is a natural $k \geq 1$ with $1 / k < ε$ , and every nonzero natural is a successor, so $k = n + 1$ for some $n \in N$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ , Every nonzero natural number is a successor, The canonical natural $ι (n) = n \cdot 1_{F}$ of a field).

[L6]

A two-element set of reals has a maximum and a minimum, each of which is one of the two elements (Maximum and minimum of a set, Every nonempty finite set of reals has a maximum and a minimum); and $[0, 1]$ is the set of reals $t$ with $0 \leq t \leq 1$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

[L7]

Every metrizable space is Hausdorff, hence $T_{1}$ and $T_{0}$ (Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not, Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not, Every Urysohn space is Hausdorff, every Hausdorff space is $T_{1}$ and hence $T_{0}$ , and every regular $T_{1}$ space is Urysohn, $T_{0}$ (Kolmogorov) and $T_{1}$ (Frechet) spaces).

[L8]

Every metric space is completely normal, hence normal (In a metric space any two separated sets have disjoint open neighbourhoods, so every metrizable space is completely normal).

Proof

technique · direct

1.1L1L2L3assume-hyp

Suppose $C \neq \emptyset$ and put $g (x) : = d (x, C)$ ; then $g$ is continuous by [L2] and [L3] with $L = 1$ .

1.2L3L4

If $C = \emptyset$ then the constant function $1$ is continuous and has zero set $\emptyset = C$ , since $1 \neq 0$ .

2.1step 1.1L1L4

Under step 1.1: $Z (g) = {x : d (x, C) = 0} = \overline{C} = C$ , the last equality because $C$ is closed.

2.2step 1.1L2

Under step 1.1: for each $n$ the set $W_{n} : = {x : d (x, C) < 1 / (n + 1)}$ is open, since for $x \in W_{n}$ and $t : = 1 / (n + 1) - d (x, C) > 0$ any $y$ with $d (x, y) < t$ has $d (y, C) \leq d (x, C) + d (x, y) < 1 / (n + 1)$ by [L2].

3.1step 2.1step 1.2

By steps 2.1 and 1.2 every closed subset of $X$ is a zero set, which is claim 1.

3.2step 2.1step 2.2L1

Under step 1.1: $C \subseteq ⋂_{n} W_{n}$ , since $d (x, C) = 0 < 1 / (n + 1)$ for $x \in C$ by [L1] and step 2.1.

3.3step 2.1step 2.2L1L5

Under step 1.1: if $x \notin C$ then $d (x, C) > 0$ by [L1] and step 2.1, so [L5] gives $n$ with $1 / (n + 1) < d (x, C)$ and hence $x \notin W_{n}$ .

3.4step 2.1L1L6

Under step 1.1 with $x_{0} \notin C$ : $r : = d (x_{0}, C) > 0$ by [L1] and step 2.1, and $f (x) : = \min {1, d (x, C) / r}$ takes values in $[0, 1]$ by [L1] and [L6].

4.1step 3.2step 3.3L4

Steps 3.2 and 3.3 give $C = ⋂_{n} W_{n}$ for nonempty closed $C$ , and $\emptyset$ is open hence a $G_{δ}$ by [L4]; this is claim 2.

4.2step 3.4L2L3L6

Under step 3.4: $∣ \min {1, u} - \min {1, v} ∣ \leq ∣ u - v ∣$ for all reals $u, v$ , since if both are at most $1$ the two sides are equal, if both exceed $1$ the left side is $0$ , and if $u \leq 1 < v$ then the left side is $1 - u$ , which is at most $v - u$ , the remaining case $v \leq 1 < u$ being the same with $u$ and $v$ exchanged; hence $∣ f (x) - f (y) ∣ \leq ∣ d (x, C) - d (y, C) ∣ / r \leq d (x, y) / r$ and $f$ is continuous by [L3] with $L = 1 / r$ .

4.3step 3.4L1L6

Under step 3.4: $f (x_{0}) = \min {1, r / r} = \min {1, 1} = 1$ , and $f (y) = \min {1, 0} = 0$ for $y \in C$ since $d (y, C) = 0$ .

5.1step 1.2step 4.2step 4.3

By steps 4.2 and 4.3, and by step 1.2 for the case $C = \emptyset$ , the space $X$ is completely regular, which is claim 3.

6.1step 4.1step 5.1L7L8

A metrizable space $Y$ is completely regular by step 5.1 applied to any inducing metric, and it is $T_{1}$ by [L7], so it is Tychonoff; it is normal by [L8] and every closed subset of it is a $G_{δ}$ by step 4.1, so it is perfectly normal.

7.1step 6.1L7L8∎

Being perfectly normal and $T_{1}$ , such a $Y$ is $T_{6}$ ; it is $T_{5}$ and $T_{4}$ by [L8] and $T_{1}$ , it is $T_{3 \frac{1}{2}}$ by step 6.1, and it is $T_{3}$ , $T_{2 \frac{1}{2}}$ , $T_{2}$ , $T_{1}$ and $T_{0}$ by the implications already proved on this page; this is claim 4.

Remarks

Claim 1 is the sharp form and claim 2 is its shadow. A zero set is always a $G_{δ}$ (Zero sets and cozero sets of continuous real-valued functions), so claim 2 follows from claim 1; it is proved separately here because the explicit presentation $⋂_{n} {x : d (x, C) < 1 / (n + 1)}$ is the one quoted later, and because it makes visible that the index runs from $n = 0$ , where the radius is $1$ .
The empty closed set is not a nuisance to be waved away. $d (x, \emptyset)$ is undefined in this library, there being no infimum of the empty set (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space), so each of the three claims is discharged separately at $C = \emptyset$ by a constant function or by openness.
What this does not prove. It says nothing about which non-metrizable spaces are perfectly normal, and it gives no metrization theorem in the other direction: exhibiting a metric is the only way a space is shown metrizable here (Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not).

In a metric space every closed set is a zero set and a $G_{δ}$ , and the distance function separates a point from a closed set, so every metrizable space is Tychonoff and perfectly normal

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

In a metric space every closed set is a zero set and a GδGδ​, and the distance function separates a point from a closed set, so every metrizable space is Tychonoff and perfectly normal

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

In a metric space every closed set is a zero set and a $G_{δ}$ , and the distance function separates a point from a closed set, so every metrizable space is Tychonoff and perfectly normal