TheoremStatement: AI-adaptedProof: AI-generatedprecheck 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 $J = \emptyset$ . 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 $G_{δ}$ , 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 · two levels

43 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

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