CorollaryStatement: AI-adaptedProof: AI-generatedPipeline-generated

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A Urysohn separation obstruction also obstructs bounded Tietze extension

Statement

Let $X$ be a normal space (Normal spaces and $T_{4}$ spaces, with the source disagreement over whether normality includes $T_{1}$ stated explicitly) and let $A, B$ be disjoint closed subsets admitting no continuous map $X \to [0, 1]$ that is zero on $A$ and one on $B$ . Then the map $g : A \cup B \to [0, 1]$ equal to zero on $A$ and one on $B$ is continuous and has no continuous extension to $X$ .

Consequently, over ZF, failure of Urysohn's lemma implies failure of bounded Tietze extension. This is an implication between existence assertions; it constructs neither a counterexample space nor a model of ZF.

Facts & Assumptions

Given: The normal space $X$ and disjoint closed $A, B$ with the stated separation obstruction.

[F1]

Closed sets in a subspace are traces of ambient closed sets (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace).

[F2]

Continuity means that preimages of open sets are open (Continuity of a map of topological spaces at a point and globally); the target is $[0, 1]$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

Proof

1.1givenF1F2construct

In the subspace $A \cup B$ , the sets $A$ and $B$ are closed by F1 and are each other's complements. Thus both are open. Define $g = 0$ on $A$ and $g = 1$ on $B$ , well-defined since the sets are disjoint. The preimage under $g$ of any open subset of $[0, 1]$ is one of $\emptyset, A, B, A \cup B$ , all open in the subspace. Hence $g$ is continuous by F2.

2.1givenstep 1.1∎

A continuous extension $G : X \to [0, 1]$ of $g$ would vanish on $A$ and be one on $B$ , contradicting the given obstruction. Therefore no such extension exists. Since $A \cup B$ is closed in $X$ , this is exactly a failure of bounded Tietze extension. Applying this construction to any normal-space witness of $\neg U R Y$ proves the final implication.

Depends on

Used by

Dependency tree · two levels

16 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

Norbert Brunner, Geordnete Läuchli Kontinuen (standard reference, not scraped)
Eleftherios Tachtsis, The Urysohn Lemma is independent of ZF + Countable Choice (standard reference, not scraped)
Eleftherios Tachtsis, Erratum to The Urysohn Lemma is independent of ZF + Countable Choice (standard reference, not scraped)