CorollaryStatement: AI-adaptedProof: AI-generatedPipeline-generatedaudited 2026-09-22

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Brunner's endpoint obstruction also refutes bounded Tietze extension

Statement

Let $L$ be the compact normal ordered continuum of Brunner's models satisfy the required choice and Urysohn obstructions with its two distinct endpoint closed sets $A$ and $B$ . The continuous map $g : A \cup B \to [0, 1]$ that is $0$ on $A$ and $1$ on $B$ has no continuous extension to $L$ . Consequently $Con (Z F)$ implies both $Con (Z F + {A C}_{ω} + failure of bounded Tietze extension)$ and $Con (Z F + B P I + failure of bounded Tietze extension)$ .

Facts & Assumptions

Given: The continuum $L$ , its two endpoint closed sets $A, B$ , and the function $g$ that is $0$ on $A$ and $1$ on $B$ .

[F1]

In $L$ every continuous real-valued function is constant, and $L$ is a compact Hausdorff, hence normal, space (Brunner's models satisfy the required choice and Urysohn obstructions, Normal spaces and $T_{4}$ spaces, with the source disagreement over whether normality includes $T_{1}$ stated explicitly, Continuity of a map of topological spaces at a point and globally).

[F2]

The subspace $A \cup B$ carries the subspace topology, in which a subset is open exactly when it is the trace of an open set of $L$ (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).

[F3]

Conditional on $Con (Z F)$ , the relative-consistency theorems of this page give, respectively, a model of $Z F + {A C}_{ω}$ and a model of $Z F + B P I$ in which some normal space has two disjoint closed sets admitting no continuous separation (Relative consistency of Countable Choice without Urysohn's lemma, Relative consistency of BPI without Urysohn's lemma).

[L1]

The interval $[0, 1]$ is a closed bounded interval of $R$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

Proof

technique · direct

1.1

The sets $A$ and $B$ are complementary closed subsets of the subspace $A \cup B$ , so each is clopen in that subspace by [F2] and [L1].

givenF2L1

2.1

By step 1.1, the map $g$ that is $0$ on the clopen set $A$ and $1$ on the clopen set $B$ is continuous on $A \cup B$ : the preimage of any subset of $[0, 1]$ is a union of some of $A$ , $B$ , both of which are open in the subspace.

step 1.1F2

3.1

Suppose $G : L \to [0, 1]$ were a continuous extension of $g$ . Then $G$ is a continuous real-valued function on $L$ , hence constant by [F1]; but $G$ equals $0$ on $A$ and $1$ on $B$ , and $A, B$ are nonempty, so no constant function can agree with $g$ .

step 2.1F1

4.1

Therefore no continuous extension of $g$ exists, which is the failure of bounded Tietze extension for the closed subspace $A \cup B$ of $L$ . For either model supplied by [F3], let $X$ be its normal-space witness and let $C, D$ be the disjoint closed sets admitting no continuous separation. The map on $C \cup D$ with values $0$ on $C$ and $1$ on $D$ is continuous by the same clopen-subspace argument as steps 1.1--2.1; any continuous extension to $X$ would separate $C$ and $D$ , contrary to their defining property. Thus each model supplied by [F3] also witnesses failure of bounded Tietze extension, giving the two displayed consistency statements.

step 1.1step 2.1step 3.1F3∎

Depends on

Used by

Dependency tree · two levels

42 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)