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

Statement refuted

The continuous function $f : (0,1] \to \mathbb{R}$ , $f(x) := 1/x$ , extends to a continuous function $F : \mathbb{R} \to \mathbb{R}$ .

This is the single witness behind FALSE: Every continuous real-valued function on a subspace of a normal space extends continuously to the whole space, presented on its own as the counterexample it is: it shows that dropping the hypothesis " $A$ closed" from 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 a minor loosening but breaks that extension statement outright, on the very space $\mathbb{R}$ where it is otherwise available.

Which statement this witness refutes, and which it does not. The corollary is the $\mathbb{R}$ -valued form, and $f$ meets every one of its hypotheses except closedness of $A$ , so it isolates that hypothesis exactly. It does not refute 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 itself with the closedness hypothesis removed: that theorem is stated for maps into a bounded interval $[a,b]$ , and $f$ is unbounded, so $f$ fails its codomain hypothesis as well. A witness violating two hypotheses cannot isolate one.

Facts & Assumptions

Given: $A := (0,1] \subseteq \mathbb{R}$ and $f : A \to \mathbb{R}$ , $f(x) := 1/x$ .

[L1]

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).

[L2]

Continuity passes to subsets of the domain (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).

[L3]

$[0,1]$ is compact (Heine-Borel by bisection: every closed bounded interval $[a,b]$ is compact); a continuous real function on a compact subset of its domain is bounded there (A continuous real function on a compact subset of $\mathbb{R}$ is bounded).

[L4]

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 $1/n < \varepsilon$ ).

Counterexample

technique · contradiction

1.1

$f$ is continuous on $A$ by [L1], with $0 \notin A$ .

givenL1

1.2

For every real $M$ there is $x \in A$ with $f(x)>M$ : for $M \le 0$ take $x:=1$ ; for $M>0$ , [L4] with $\varepsilon := 1/(M+1)$ gives a natural $n \ge 1$ with $1/n < 1/(M+1)$ , hence $n>M$ , and $x := 1/n \in (0,1]=A$ has $f(x)=n>M$ .

givenL4algebrachoose

1.3

Suppose, toward a contradiction, that a continuous $F : \mathbb{R} \to \mathbb{R}$ extends $f$ .

assume-contra

2.1

Under step 1.3: $F|_{[0,1]}$ is continuous by [L2]; by [L3], $[0,1]$ is compact and $F|_{[0,1]}$ is therefore bounded: fix real $M_0 \ge 0$ with $|F(x)| \le M_0$ for every $x \in [0,1]$ .

step 1.3L2L3choose

3.1

Under step 1.3: for $x \in A \subseteq [0,1]$ , $F(x)=f(x)$ , so $f(x) \le M_0$ for every $x \in A$ by step 2.1; but step 1.2 with $M:=M_0$ gives $x_0 \in A$ with $f(x_0)>M_0$ , a contradiction.

step 1.3step 2.1step 1.2discharge-contradiction∎

Remarks

The only hypothesis of the $\mathbb{R}$ -valued extension statement that fails here is closedness of $A$ . $\mathbb{R}$ is normal and $f$ is continuous on $A$ ; the closure of $A$ in $\mathbb{R}$ is $[0,1]$ , and it is exactly the missing point $0$ where $f$ has nowhere finite to go. Against 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 itself the witness fails a second hypothesis, since that theorem takes values in a bounded interval and $f$ does not, which is why the statement refuted above is framed against 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.
The obstruction is boundedness, not the existence of a limit. Step 1.2 shows $f$ is unbounded on every neighbourhood of the missing point directly from the reciprocal's growth, with no appeal to $\lim_{x \to 0^+} f(x)$ failing to exist as a real number.

Depends on

Used by

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Sources

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

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