False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passverified 2026-09-09 (gpt-6-astra)

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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.

The witness isolates the closed-subspace hypothesis in the real-valued form, clause 1 of 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. Even under that corollary's dependent-choice assumption, dropping closedness permits a continuous map with no continuous extension. The nonextension proof below uses no choice principle. The bounded-range version 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 is a different comparison: the reciprocal violates its bounded-range hypothesis as well as closedness.

Facts & Assumptions

Given: $R$ with its usual topology, the subspace $A : = (0, 1] \subseteq R$ (Intervals of $R$ : 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 $f : A \to R$ , $f (x) : = 1 / x$ .

[L1]

$R$ 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 ${x \in R : x \neq 0} \supseteq A$ .

[L3]

Continuity passes to subsets of the domain: if $B \subseteq C \subseteq R$ and $g : C \to R$ is continuous, then $g ∣_{B}$ is continuous (Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ 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 $M \geq 0$ with $∣ g (x) ∣ \leq M$ for every $x \in K$ (A continuous real function on a compact subset of $R$ is bounded).

[L6]

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

Refutation

technique · contradiction

1.1

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

givenL1L2

1.2

For every real $M$ , there is $x \in A$ with $f (x) > M$ : if $M \leq 0$ , take $x : = 1$ , so $f (1) = 1 > 0 \geq M$ ; if $M > 0$ , [L6] applied to $ε : = 1 / (M + 1) > 0$ gives a natural $n \geq 1$ 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 $F : R \to R$ 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 $R \supseteq [0, 1]$ .

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 $M_{0} \geq 0$ with $∣ F (x) ∣ \leq M_{0}$ 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 $f (x) \leq ∣ F (x) ∣ \leq M_{0}$ 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 $f (x_{0}) \leq M_{0}$ .

step 1.3step 3.1step 1.2discharge-contradiction∎

Remarks

Closedness is isolated in the real-valued form. $R$ is normal and $f$ is continuous on $A$ (step 1.1), but $A$ is not closed: its closure is $[0, 1]$ . Thus, even assuming dependent choice, the witness meets every hypothesis of clause 1 of 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 except closedness. It does not isolate closedness 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, since step 1.2 also rules out any bounded interval containing its range.
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 · two levels

69 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)
Archimedean property (Wikipedia) (standard reference, not scraped)