TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29

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In a metric space any two separated sets have disjoint open neighbourhoods, so every metrizable space is completely 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 let $A, B \subseteq X$ be separated (Separated sets: $\overline{A} \cap B = A \cap \overline{B} = \emptyset$ ). Then there are disjoint open sets $U \supseteq A$ and $V \supseteq B$ .

Consequently every metrizable space (Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not) is completely normal, and hence normal (Completely normal ( $T_{5}$ ) and perfectly normal ( $T_{6}$ ) spaces, Normal spaces and $T_{4}$ spaces, with the source disagreement over whether normality includes $T_{1}$ stated explicitly).

No choice principle is used. The two open sets are unions indexed by the points of $A$ and of $B$ , and the radius attached to a point is the number $d (a, B) / 2$ , which is determined by $a$ , by $B$ and by $d$ ; nothing is selected.

Facts & Assumptions

Given: A metric space $(X, d)$ and separated sets $A, B \subseteq X$ , so that $\overline{A} \cap B = A \cap \overline{B} = \emptyset$ , with closures taken in the metric topology.

[A1]

$A$ and $B$ are separated: $\overline{A} \cap B = \emptyset$ and $A \cap \overline{B} = \emptyset$ (Separated sets: $\overline{A} \cap B = A \cap \overline{B} = \emptyset$ ).

[L1]

For nonempty $S \subseteq X$ and $x \in X$ the distance $d (x, S) = \inf {d (x, s) : s \in S}$ exists in $R$ , is a lower bound of that set, and satisfies $d (x, S) \geq 0$ (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space, Greatest lower bound (infimum), Every nonempty set bounded below has an infimum, Nonnegativity of a metric is a consequence of the other axioms, not an axiom).

[L2]

For nonempty $S \subseteq X$ , $\overline{S} = {x \in X : d (x, S) = 0}$ (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).

[L3]

Open balls are open and an arbitrary union of open sets is open; $\emptyset$ and $X$ are open (Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed, 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).

[L4]

$x \in B (x, r)$ for every $r > 0$ , and $y \in B (x, r)$ means $d (x, y) < r$ (Open ball, closed ball and sphere in a metric space).

[L5]

The triangle inequality $d (p, q) \leq d (p, x) + d (x, q)$ and symmetry $d (p, q) = d (q, p)$ (Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric).

[L6]

A two-element set of reals has a maximum, which is one of the two and is at least the other (Maximum and minimum of a set).

Proof

technique · direct

1.1

If $A = \emptyset$ then $U : = \emptyset$ and $V : = X$ are disjoint open sets with $A \subseteq U$ and $B \subseteq V$ ; if $B = \emptyset$ then $U : = X$ and $V : = \emptyset$ do the same.

L3construct

1.2

Assume from here that $A$ and $B$ are both nonempty, so that $d (x, A)$ and $d (x, B)$ are defined for every $x \in X$ .

L1assume-hyp

2.1

For $a \in A$ : $a \notin \overline{B}$ by [A1], so $d (a, B) \neq 0$ by [L2], and $d (a, B) \geq 0$ by [L1]; hence $r_{a} : = d (a, B) / 2 > 0$ . Symmetrically $s_{b} : = d (b, A) / 2 > 0$ for $b \in B$ .

step 1.2A1L1L2

3.1

Define $U : = ⋃_{a \in A} B (a, r_{a})$ and $V : = ⋃_{b \in B} B (b, s_{b})$ ; both are open by [L3], and $A \subseteq U$ and $B \subseteq V$ by [L4].

step 2.1L3L4construct

4.1

Suppose $x \in U \cap V$ ; then there are $a \in A$ and $b \in B$ with $d (a, x) < r_{a}$ and $d (b, x) < s_{b}$ .

step 3.1L4assume-hyp

5.1

Under step 4.1: $d (a, b) \leq d (a, x) + d (x, b) < r_{a} + s_{b}$ , using symmetry for $d (x, b) = d (b, x)$ .

step 4.1L5

5.2

Under step 4.1: $r_{a} + s_{b} \leq 2 \max {r_{a}, s_{b}} = \max {d (a, B), d (b, A)}$ , by [L6] and the definitions of $r_{a}$ and $s_{b}$ .

step 2.1step 4.1L6

5.3

$d (a, B) \leq d (a, b)$ , since $b \in B$ makes $d (a, b)$ a member of the set whose infimum is $d (a, B)$ ; and $d (b, A) \leq d (b, a) = d (a, b)$ for the same reason with the roles exchanged.

step 4.1L1L5

6.1

By steps 5.1, 5.2 and 5.3, $d (a, b) < \max {d (a, B), d (b, A)} \leq d (a, b)$ , which is impossible; so no such $x$ exists and $U \cap V = \emptyset$ .

step 5.1step 5.2step 5.3

7.1

By steps 1.1, 3.1 and 6.1 the separated pair $A, B$ has disjoint open supersets in every case.

step 1.1step 3.1step 6.1

8.1

If $(Y, T)$ is metrizable, fix a metric $d$ inducing $T$ ; separation of two subsets is a statement about the closure operator, and the topological closure of a metrizable space is the metric closure of any inducing metric, so step 7.1 applies verbatim and $Y$ is completely normal, hence normal.

step 7.1L2∎

Remarks

The halving is what makes the balls miss each other. Radii $d (a, B)$ and $d (b, A)$ without the factor $2$ would not do: two balls of those radii can meet, and the triangle inequality then gives no contradiction. With the halving the sum of the two radii is at most the larger of the two distances, which is at most $d (a, b)$ .
Separated, not merely disjoint, is exactly the right hypothesis. For disjoint sets the radii can fail to be positive: in $R$ the disjoint sets $(0, 1)$ and $[1, 2)$ have $d (1, (0, 1)) = 0$ , and indeed they are not separated. What the hypothesis buys is positivity of every radius, and nothing else.
The corresponding statement for $R$ needs no new proof. $R$ with its usual topology is metrizable by the usual metric (Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not), so it is completely normal, and so is every $R^{n}$ and every subspace of a metrizable space.

Depends on

Used by

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Sources

Normal space (Wikipedia) (standard reference, not scraped)
Metric space (Wikipedia) (standard reference, not scraped)
J. Munkres, Topology, 2nd ed., §32 (standard reference, not scraped)
R. Gardner, Introduction to Topology, notes on Munkres Section 32: Normal Spaces (East Tennessee State University) (standard reference, not scraped)