CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)verified 2026-07-29 (claude-fable-5)

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For continuous $f, g : Z \to Y$ with $Y$ Hausdorff the agreement set ${z \in Z : f (z) = g (z)}$ is closed in $Z$

Statement

Let $Z$ be a topological space, let $Y$ be a Hausdorff space (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not) and let $f, g : Z \to Y$ be continuous (Continuity of a map of topological spaces at a point and globally). Then the agreement set

$E (f, g) : = {z \in Z : f (z) = g (z)}$

is closed in $Z$ .

No hypothesis is placed on $Z$ : the separation hypothesis is on the codomain alone, and it is not decoration. Let $Y_{0} = {a, b}$ with $a \neq b$ carry the indiscrete topology (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies), which is not Hausdorff. Every function $Z \to Y_{0}$ is continuous, the only preimages to check being those of $\emptyset$ and $Y_{0}$ , namely $\emptyset$ and $Z$ . So for any subset $S \subseteq Z$ the constant map $f_{0} \equiv a$ and the map $g_{0}$ taking the value $a$ on $S$ and $b$ off $S$ are continuous with $E (f_{0}, g_{0}) = S$ , closed or not.

Facts & Assumptions

Given: Topological spaces $Z$ and $Y$ with $Y$ Hausdorff, continuous maps $f, g : Z \to Y$ , and the product $Y \times Y$ with the product topology.

[A1]

$E (f, g) = ⟨ f, g ⟩^{- 1} [Δ_{Y}]$ , where $⟨ f, g ⟩ : Z \to Y \times Y$ is the pairing and $Δ_{Y}$ the diagonal (The diagonal $Δ_{X} \subseteq X \times X$ , the diagonal map $δ_{X}$ , and the pairing $⟨ f, g ⟩$ of two maps).

[L1]

The pairing $⟨ f, g ⟩$ is continuous whenever $f$ and $g$ are ( $δ_{X}$ is a topological embedding of $X$ onto $Δ_{X}$ , and $⟨ f, g ⟩$ is continuous whenever $f$ and $g$ are, claim 1).

[L2]

$Δ_{Y}$ is closed in $Y \times Y$ , since $Y$ is Hausdorff (A space is Hausdorff if and only if its diagonal is closed in the square carrying the product topology, Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not).

[L3]

A map is continuous if and only if the preimage of every closed set is closed (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 $f (\overline{A}) \subseteq \overline{f (A)}$ , clauses (a) and (c), Continuity of a map of topological spaces at a point and globally).

Proof

technique · direct

1.1

$⟨ f, g ⟩ : Z \to Y \times Y$ is continuous.

1.2

$Δ_{Y}$ is closed in $Y \times Y$ .

2.1

$E (f, g) = ⟨ f, g ⟩^{- 1} [Δ_{Y}]$ is the preimage of a closed set under a continuous map, hence closed in $Z$ .

step 1.1step 1.2A1L3∎

Remarks

Why the diagonal criterion is the right tool here. The condition " $f (z) = g (z)$ " is a condition on the pair of values, so it becomes a membership condition once the two maps are packaged into one map into the square; the criterion then converts the separation hypothesis on $Y$ into the closedness of the set that condition names. Nothing is proved twice: the whole content is A space is Hausdorff if and only if its diagonal is closed in the square carrying the product topology together with the preimage identity of The diagonal $Δ_{X} \subseteq X \times X$ , the diagonal map $δ_{X}$ , and the pairing $⟨ f, g ⟩$ of two maps.
Both hypotheses are used, and only these. Continuity of $f$ and $g$ enters only through [L1], and the Hausdorff condition only through [L2]. In particular no countability, compactness or separation hypothesis on $Z$ appears anywhere in the argument.
The complement is what the statement is often used for. $Z ∖ E (f, g)$ is open, so if $f$ and $g$ differ at a point they differ throughout some open neighbourhood of it. Equivalently, $E (f, g)$ contains the closure of every subset of $Z$ on which $f$ and $g$ agree, which is the form in which a statement about a dense set is obtained from this one.

Depends on

Used by

Dependency tree · two levels

28 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Hausdorff space (Wikipedia) (standard reference, not scraped)
Continuous function (Wikipedia) (standard reference, not scraped)
General Topology Notes (UC Riverside) (standard reference, not scraped)

For continuous f,g:Z→Yf,g:Z→Y with YY Hausdorff the agreement set {z∈Z:f(z)=g(z)}{z∈Z:f(z)=g(z)} is closed in ZZ