CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)verified 2026-07-29 (claude-fable-5)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

In the $K$ -topology on $R$ the closed set $K \cup {0}$ carries a continuous two-valued function with no continuous extension

Statement refuted

The subspace $A : = K \cup {0} \subseteq (R, T_{K})$ of the $K$ -topology (The $K$ -topology on $R$ , generated by the open intervals together with their complements of $K = {1 / (n + 1) : n \in N}$ , is $T_{1}$ and Hausdorff but not regular) is closed, and the continuous $k : A \to [0, 1]$ with $k \equiv 0$ on $K$ and $k (0) : = 1$ extends to a continuous $F : R \to [0, 1]$ .

This is false, and the failure is not a failure of closedness: $A$ is closed. What fails is normality of the ambient space — $(R, T_{K})$ is $T_{1}$ but not normal — in contrast with the companion witness on this page that instead varies the closedness hypothesis of Tietze's theorem while keeping the ambient space normal.

Facts & Assumptions

Given: $(R, T_{K})$ the $K$ -topology with basis $B_{K} = {(a, b) : a < b} \cup {(a, b) ∖ K : a < b}$ and $K = {1 / (n + 1) : n \in N}$ (The $K$ -topology on $R$ , generated by the open intervals together with their complements of $K = {1 / (n + 1) : n \in N}$ , is $T_{1}$ and Hausdorff but not regular); $A : = K \cup {0}$ ; and $k : A \to [0, 1]$ , $k \equiv 0$ on $K$ , $k (0) : = 1$ .

[L1]

$(R, T_{K})$ is $T_{1}$ ; $K$ is closed in $T_{K}$ ; and the point $0$ and the closed set $K$ admit no disjoint open neighbourhoods, i.e. $(R, T_{K})$ is not regular (The $K$ -topology on $R$ , generated by the open intervals together with their complements of $K = {1 / (n + 1) : n \in N}$ , is $T_{1}$ and Hausdorff but not regular, clauses 2–4).

[L2]

Every singleton is closed in a $T_{1}$ space (A space is $T_{1}$ if and only if every singleton is closed, if and only if every finite subset is closed, if and only if its topology contains the cofinite topology).

[L3]

For $a < b$ , $(a, b) ∖ K \in B_{K}$ , hence open in $T_{K}$ (The $K$ -topology on $R$ , generated by the open intervals together with their complements of $K = {1 / (n + 1) : n \in N}$ , is $T_{1}$ and Hausdorff but not regular).

[L4]

Preimages of open sets under a continuous map are open (For a map of spaces the following agree: continuity at every point, preimages of open sets open, preimages of closed sets closed, preimages of subbasic open sets open, and $f (\overline{A}) \subseteq \overline{f (A)}$ , clause (b)).

Counterexample

technique · contradiction

1.1

$0 \notin K$ : every $1 / (n + 1)$ , $n \in N$ , is positive, and $0$ is not. By [L2] (using [L1], $T_{1}$ ), ${0}$ is closed; with $K$ closed by [L1], $A = K \cup {0}$ is closed, a union of two closed sets.

givenL1L2algebra

1.2

${0}$ is open in the subspace $A$ : $(- 1, 1) ∖ K$ is open in $T_{K}$ by [L3], and its trace on $A$ is $((- 1, 1) ∖ K) \cap (K \cup {0}) = {0}$ , since it excludes every point of $K$ and contains $0$ .

givenL3algebra

1.3

Suppose, toward a contradiction, that a continuous $F : R \to [0, 1]$ exists with $F ∣_{A} = k$ .

assume-contra

2.1

$K$ is open in the subspace $A$ : $R ∖ {0}$ is open in $T_{K}$ by step 1.1 ( ${0}$ closed), and its trace on $A$ is $(R ∖ {0}) \cap (K \cup {0}) = K$ , since $0 \notin K$ .

step 1.1algebra

2.2

Under step 1.3: $W_{1} : = F^{- 1} [(- \frac{1}{2}, \frac{1}{2})]$ and $W_{2} : = F^{- 1} [(\frac{1}{2}, \frac{3}{2})]$ are open in $T_{K}$ by [L4]. $K \subseteq W_{1}$ , since $F \equiv k \equiv 0 \in (- \frac{1}{2}, \frac{1}{2})$ on $K \subseteq A$ ; $0 \in W_{2}$ , since $F (0) = k (0) = 1 \in (\frac{1}{2}, \frac{3}{2})$ ; and $W_{1} \cap W_{2} = \emptyset$ , the target intervals $(- \frac{1}{2}, \frac{1}{2})$ and $(\frac{1}{2}, \frac{3}{2})$ being disjoint.

step 1.3L4algebra

3.1

$k$ is continuous on $A$ : for open $V \subseteq [0, 1]$ , $k^{- 1} [V]$ is $A$ if $0, 1 \in V$ ; $K$ if $0 \in V, 1 \notin V$ (open in $A$ by step 2.1); ${0}$ if $1 \in V, 0 \notin V$ (open in $A$ by step 1.2); or $\emptyset$ otherwise; in every case open in $A$ .

step 1.2step 2.1

4.1

Step 2.2 exhibits disjoint open $W_{2} ∋ 0$ and $W_{1} \supseteq K$ , contradicting [L1]: the point $0$ and the closed set $K$ admit no disjoint open neighbourhoods.

step 2.2L1discharge-contradiction∎

Remarks

$(R, T_{K})$ is not normal, and this is why Tietze's theorem does not apply here. It is $T_{1}$ (step 1.1's citation of [L1]) but not regular (used directly in step 4.1); by A normal $T_{1}$ space is regular, hence $T_{3}$ , hence Urysohn, Hausdorff, $T_{1}$ and $T_{0}$ , a normal $T_{1}$ space is regular, so a $T_{1}$ space that is not regular cannot be normal. Tietze's extension theorem, under dependent choice: a continuous map from a closed subspace of a normal space into $[a, b]$ extends continuously to the whole space, and this property characterises normality requires normality of the ambient space, and that hypothesis is exactly what fails.
Closedness of $A$ is not in question. $A$ is closed by step 1.1, and $k$ is continuous on $A$ by step 3.1; every hypothesis of Tietze's extension theorem, under dependent choice: a continuous map from a closed subspace of a normal space into $[a, b]$ extends continuously to the whole space, and this property characterises normality except normality of $R$ with the $T_{K}$ topology holds here.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

47 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

Tietze extension theorem (Wikipedia) (standard reference, not scraped)
K-topology (Wikipedia) (standard reference, not scraped)

In the KK-topology on RR the closed set K∪{0}K∪{0} carries a continuous two-valued function with no continuous extension