TheoremStatement: Literature-sourcedProof: Literature-sourcedPipeline-generatedaudited 2026-09-22

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Relative consistency of Countable Choice without Urysohn's lemma

Statement

If $Z F$ is consistent, then $Z F + {A C}_{ω} + failure of Urysohn’s lemma$ is consistent: there is a model of $Z F$ with countable choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )) in which some normal space has two disjoint closed sets admitting no continuous separation (Normal spaces and $T_{4}$ spaces, with the source disagreement over whether normality includes $T_{1}$ stated explicitly).

Facts & Assumptions

Given: The assumed consistency of $Z F$ .

[F1]

Tachtsis's cited theorem, read together with its published erratum, proves the exact external implication $Con (Z F) ⟹ Con (Z F + {A C}_{ω} + \neg U R Y) .$ Here $U R Y$ is the usual Urysohn separation assertion for disjoint closed subsets of a normal space. This item records that published relative-consistency theorem; it does not reconstruct the permutation-model and transfer argument.

[F2]

By definition, $\neg U R Y$ supplies a normal space $X$ and disjoint closed subsets $A, B \subseteq X$ for which there is no continuous $f : X \to [0, 1]$ satisfying $A \subseteq f^{- 1} ({0}) and B \subseteq f^{- 1} ({1}) .$ Equivalently, such an $f$ would have $f (a) = 0$ for every $a \in A$ and $f (b) = 1$ for every $b \in B$ (Normal spaces and $T_{4}$ spaces, with the source disagreement over whether normality includes $T_{1}$ stated explicitly). Equality of the images with the endpoint singletons is not the definition: it is too strong when either closed set is empty.

Proof technique: direct.

Proof

1.1

Assume $Con (Z F)$ . The published relative-consistency theorem [F1], with its erratum included in the cited interface, yields a model of $Z F + {A C}_{ω} + \neg U R Y$ .

givenF1

2.1

In that model ${A C}_{ω}$ is precisely countable choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )), while [F2] expands $\neg U R Y$ as a normal space with two disjoint closed sets admitting no continuous Urysohn separator. Thus the model has exactly the two properties asserted in the Statement.

step 1.1F2

3.1

Therefore $Con (Z F) ⟹ Con (Z F + {A C}_{ω} + failure of Urysohn’s lemma)$ , as claimed. This proof depends on the corrected published theorem itself and makes no unsupported Pincus transfer of countable choice alone.

step 2.1F1∎

Depends on

Used by

Brunner's endpoint obstruction also refutes bounded Tietze extension Corollary
If ZF is consistent, DMC is not provable in ZF Corollary
If ZF is consistent, ZF does not prove Urysohn's lemma Corollary
Choice ledger for Baire, Urysohn, Stone, and Tychonoff Remark
The converse from Urysohn's lemma to DMC is open Remark

Dependency tree · two levels

12 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

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)