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LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passverified 2026-08-09 (gpt-5.6-terra-codex-subscription)
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The topology of a nonempty completely regular space is induced by the gauge of its continuous [0,1][0,1]-valued pseudometrics

Statement

The topology of a nonempty completely regular space is induced by the gauge of pseudometrics pf(x,y)=f(x)f(y)p_f(x,y)=|f(x)-f(y)|, where f:X[0,1]f:X\to[0,1] ranges over continuous maps.

Facts & Assumptions

Given: A nonempty completely regular space XX.

[L1]

Complete regularity separates a point from a closed set by a continuous [0,1][0,1]-valued function (Completely regular spaces and Tychonoff (T312T_{3\frac{1}{2}}) spaces, Intervals of R\mathbb{R}: the nine order-convex forms, nondegeneracy, and length).

[L2]

Such functions are continuous in the neighbourhood sense (Continuity of a map of topological spaces at a point and globally).

[L3]

A gauge generates a uniformity from finite simultaneous pseudometric balls (A gauge of pseudometrics and, on a nonempty set, the uniformity it generates).

[L4]

Absolute value is nonnegative, vanishes only at zero and is even (Basic properties of the absolute value), and it satisfies u+vu+v|u+v|\le |u|+|v| (The triangle inequality).

Proof

technique · constructive
1.1

For each continuous f:X[0,1]f:X\to[0,1], direct substitution in [L4] shows that pf(x,y)=f(x)f(y)p_f(x,y)=|f(x)-f(y)| is nonnegative, symmetric, zero on the diagonal and satisfies the triangle inequality, so it is a pseudometric; its balls about xx are original-open by [L2].

L2L4construct
1.2

Conversely, if xUx\in U is original-open, apply [L1] to the closed set XUX\setminus U to obtain ff with f(x)=1f(x)=1 and f[XU]={0}f[X\setminus U]=\{0\}; then the pfp_f-ball of radius 1/21/2 about xx lies in UU.

L1L3choose
2.1

Hence every gauge-open set is original-open.

step 1.1L3
3.1

Thus original-open and gauge-open sets contain one another, so the two topologies agree.

step 2.1step 1.2discharge-construct

Depends on

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