TheoremStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02

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A space is Tychonoff if and only if it embeds in a cube $[0,1]^J$

Statement

A topological space $X$ is Tychonoff (Completely regular spaces and Tychonoff ( $T_{3\frac{1}{2}}$ ) spaces) if and only if there are a set $J$ and a topological embedding of $X$ into the cube $[0,1]^J$ . The assertion includes $J=\varnothing$ . More specifically, when $X$ is Tychonoff, its full evaluation map $e:X\to[0,1]^{C(X,[0,1])}$ is such an embedding.

Facts & Assumptions

Given: A topological space $X$ .

[L1]

A completely regular space separates every point from every disjoint closed set by a continuous map to $[0,1]$ , and a Tychonoff space is completely regular and $T_1$ (Completely regular spaces and Tychonoff ( $T_{3\frac{1}{2}}$ ) spaces).

[L2]

Products preserve complete regularity and $T_1$ , and subspaces inherit each property (Arbitrary products of completely regular spaces are completely regular, Arbitrary products preserve $T_0$ , $T_1$ , and Hausdorffness, Complete regularity is hereditary, without a hidden $T_1$ hypothesis, $T_0$ , $T_1$ , and Hausdorffness are hereditary).

[L3]

Every metric space is Tychonoff (In a metric space every closed set is a zero set and a $G_\delta$ , and the distance function separates a point from a closed set, so every metrizable space is Tychonoff and perfectly normal).

[L4]

A point–closed-set separating family has an evaluation map that is a topological embedding (The evaluation map of a point–closed-set separating family is a topological embedding).

Proof

technique · direct

1.1

If $X$ is Tychonoff, let $J=C(X,[0,1])$ . By [L1], complete regularity separates a point from a closed set and $T_1$ makes singletons closed, so this full family separates both points and points from closed sets. Hence [L4] embeds $X$ in $[0,1]^J$ .

L1L4

1.2

Conversely, suppose $X$ embeds in $[0,1]^J$ . The interval is a metric space and hence Tychonoff by [L3], so [L2] makes the cube completely regular and $T_1$ , and then makes its subspace $X$ completely regular and $T_1$ .

L2L3

2.1

The two implications are steps 1.1 and 1.2, so the equivalence holds.

step 1.1step 1.2∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 107 results over 18 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

E. Moorhouse, The Stone–Čech Compactification (standard reference, not scraped)

A space is Tychonoff if and only if it embeds in a cube [0,1]J[0,1]^J[0,1]J