CounterexampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)verified 2026-07-29 (claude-fable-5)

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

Statement refuted

The subspace $A := K \cup \{0\} \subseteq (\mathbb{R}, \mathcal{T}_K)$ of the $K$ -topology (The $K$ -topology on $\mathbb{R}$ , generated by the open intervals together with their complements of $K = \{1/(n+1) : n \in \mathbb{N}\}$ , 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 $F : \mathbb{R} \to [0,1]$ .

This is false, and the failure is not a failure of closedness: $A$ is closed. What fails is normality of the ambient space — $(\mathbb{R}, \mathcal{T}_K)$ 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: $(\mathbb{R},\mathcal{T}_K)$ the $K$ -topology with basis $\mathcal{B}_K = \{(a,b) : a<b\} \cup \{(a,b)\setminus K : a<b\}$ and $K = \{1/(n+1) : n \in \mathbb{N}\}$ (The $K$ -topology on $\mathbb{R}$ , generated by the open intervals together with their complements of $K = \{1/(n+1) : n \in \mathbb{N}\}$ , 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]

$(\mathbb{R},\mathcal{T}_K)$ is $T_1$ ; $K$ is closed in $\mathcal{T}_K$ ; and the point $0$ and the closed set $K$ admit no disjoint open neighbourhoods, i.e. $(\mathbb{R},\mathcal{T}_K)$ is not regular (The $K$ -topology on $\mathbb{R}$ , generated by the open intervals together with their complements of $K = \{1/(n+1) : n \in \mathbb{N}\}$ , 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 $a<b$ , $(a,b)\setminus K \in \mathcal{B}_K$ , hence open in $\mathcal{T}_K$ (The $K$ -topology on $\mathbb{R}$ , generated by the open intervals together with their complements of $K = \{1/(n+1) : n \in \mathbb{N}\}$ , 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 $1/(n+1)$ , $n \in \mathbb{N}$ , 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 $A$ : $(-1,1)\setminus K$ is open in $\mathcal{T}_K$ by [L3], and its trace on $A$ is $\big((-1,1)\setminus K\big) \cap (K \cup \{0\}) = \{0\}$ , since it excludes every point of $K$ and contains $0$ .

givenL3algebra

1.3

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

assume-contra

2.1

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

step 1.1algebra

2.2

Under step 1.3: $W_1 := F^{-1}\big[(-\tfrac12,\tfrac12)\big]$ and $W_2 := F^{-1}\big[(\tfrac12,\tfrac32)\big]$ are open in $\mathcal{T}_K$ by [L4]. $K \subseteq W_1$ , since $F \equiv k \equiv 0 \in (-\tfrac12,\tfrac12)$ on $K \subseteq A$ ; $0 \in W_2$ , since $F(0) = k(0) = 1 \in (\tfrac12,\tfrac32)$ ; and $W_1 \cap W_2 = \varnothing$ , the target intervals $(-\tfrac12,\tfrac12)$ and $(\tfrac12,\tfrac32)$ 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 $\varnothing$ otherwise; in every case open in $A$ .

step 1.2step 2.1

4.1

Step 2.2 exhibits disjoint open $W_2 \ni 0$ 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

$(\mathbb{R},\mathcal{T}_K)$ 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 $\mathbb{R}$ with the $\mathcal{T}_K$ topology holds here.

Depends on

Used by

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Sources

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

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