False statementConstruction: AI-adaptedVerification: 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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FALSE: Every continuous real-valued function on a subspace of a normal space extends continuously to the whole space

Statement

FALSE. Every continuous real-valued function on a subspace of a normal space extends continuously to the whole space.

This shows that the hypothesis " $A$ closed" in 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 and Under dependent choice, a continuous real-valued map on a closed subspace of a normal space extends to the whole space, and a map into an open interval extends into that same open interval is not decoration: the witness below is a continuous function on a subspace of a normal space that has no continuous extension at all, and the only hypothesis it fails is closedness of the subspace.

Facts & Assumptions

Given: $\mathbb{R}$ with its usual topology, the subspace $A := (0,1] \subseteq \mathbb{R}$ (Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length, 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), and $f : A \to \mathbb{R}$ , $f(x) := 1/x$ .

[L1]

$\mathbb{R}$ is normal, being metrizable (In a metric space any two separated sets have disjoint open neighbourhoods, so every metrizable space is completely normal).

[L2]

Quotients of continuous real functions with nonvanishing denominator are continuous (Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function, clause 4); in particular $x \mapsto 1/x$ is continuous on $\{x \in \mathbb{R} : x \ne 0\} \supseteq A$ .

[L3]

Continuity passes to subsets of the domain: if $B \subseteq C \subseteq \mathbb{R}$ and $g : C \to \mathbb{R}$ is continuous, then $g|_B$ is continuous (Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$ : the $\varepsilon$ - $\delta$ condition, its agreement with $\lim_{x \to c} f(x) = f(c)$ at a limit point, and continuity at an isolated point).

[L4]

$[0,1]$ is compact (Heine-Borel by bisection: every closed bounded interval $[a,b]$ is compact).

[L5]

A continuous real function on a compact subset $K$ of its domain is bounded on $K$ : there is real $M \ge 0$ with $|g(x)| \le M$ for every $x \in K$ (A continuous real function on a compact subset of $\mathbb{R}$ is bounded).

[L6]

For every real $\varepsilon>0$ there is a natural $n \ge 1$ with $1/n < \varepsilon$ (For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ ).

Refutation

technique · contradiction

1.1

$f$ is continuous on $A$ , by [L2] with $0 \notin A$ ; and $\mathbb{R}$ is normal, by [L1].

givenL1L2

1.2

For every real $M$ , there is $x \in A$ with $f(x) > M$ : if $M \le 0$ , take $x:=1$ , so $f(1)=1>0\ge M$ ; if $M>0$ , [L6] applied to $\varepsilon := 1/(M+1) > 0$ gives a natural $n \ge 1$ with $1/n < 1/(M+1)$ , hence $n > M+1 > M$ ; taking $x := 1/n \in (0,1] = A$ gives $f(x) = 1/x = n > M$ .

givenL6algebrachoose

1.3

Suppose, toward a contradiction, that a continuous $F : \mathbb{R} \to \mathbb{R}$ exists with $F|_A = f$ .

assume-contra

1.4

$[0,1]$ is compact, by [L4].

2.1

Under step 1.3: $F|_{[0,1]}$ is continuous, by [L3] applied to $F$ on $\mathbb{R} \supseteq [0,1]$ .

step 1.3L3

3.1

Under step 1.3: by [L5] applied to $F|_{[0,1]}$ (step 2.1) and $K := [0,1]$ (step 1.4), fix a real $M_0 \ge 0$ with $|F(x)| \le M_0$ for every $x \in [0,1]$ .

step 2.1step 1.4L5choose

4.1

Under step 1.3: for $x \in A$ , $F(x) = f(x)$ (step 1.3) and $x \in [0,1]$ , so $f(x) \le |F(x)| \le M_0$ by step 3.1; but step 1.2 applied with $M := M_0$ gives $x_0 \in A$ with $f(x_0) > M_0$ , contradicting $f(x_0) \le M_0$ .

step 1.3step 3.1step 1.2discharge-contradiction∎

Remarks

No property but closedness fails. $\mathbb{R}$ is normal (step 1.1), $f$ is continuous on $A$ (step 1.1), and the target is all of $\mathbb{R}$ , so 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 holds except that $A$ is not closed in $\mathbb{R}$ — its closure is $[0,1]$ , one point larger.
The obstruction is unboundedness near the missing point, not discontinuity. $f$ itself is continuous at every point of its own domain $A$ ; nothing about $f$ is badly behaved on $A$ . What blocks an extension is that $f$ has no finite value it could sensibly take at the boundary point $0 \notin A$ , and step 1.2 makes that failure of boundedness explicit rather than appealing to a limit that does not exist.

Depends on

Used by

The reciprocal on (0,1] is continuous and extends to no continuous function on ℝ, so closedness of the subspace is not decoration in the ℝ-valued Tietze extension Counterexample

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 163 results over 20 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Tietze extension theorem (Wikipedia) (standard reference, not scraped)
Archimedean property (Wikipedia) (standard reference, not scraped)