CounterexampleConstruction: Literature-sourcedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck 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.

Refuted: a function into a Hausdorff space whose graph is closed is continuous. The function equal to $1/x$ off $0$ and to $0$ at $0$ has a closed graph, is discontinuous at $0$ alone, and has a Hausdorff codomain

Statement refuted

False claim: if $X$ is a topological space, $Y$ is Hausdorff (Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not) and $f : X \to Y$ has graph closed in $X \times Y$ with the product topology (The product set $\prod_{i \in I} X_i$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space), then $f$ is continuous (Continuity of a map of topological spaces at a point and globally).

This is the sharpening of FALSE: every function between topological spaces whose graph is closed in the product is continuous that adds to the codomain the hypothesis under which the other half of the closed-graph criterion holds. It is still false, and the same witness refutes it:

$f : \mathbb{R} \to \mathbb{R}, \qquad f(x) = \frac{1}{x} \ \ (x \ne 0), \qquad f(0) = 0 ,$

with $\mathbb{R}$ carrying its usual topology, which is metrizable and hence Hausdorff. Its graph is closed in $\mathbb{R}^2$ , it is continuous at every $c \ne 0$ , and it is not continuous at $0$ ; so its set of discontinuities is exactly $\{0\}$ .

What the criterion of A map into a compact space whose graph is closed is continuous; so for a compact Hausdorff codomain, continuity and closedness of the graph are equivalent asks of the codomain in the direction "closed graph implies continuous" is compactness, and the Hausdorff condition contributes nothing there.

Facts & Assumptions

Given: $\mathbb{R}$ with its usual topology, $\mathbb{R}^2 = \mathbb{R} \times \mathbb{R}$ with the product topology, and the function $f$ above with graph $G_f$ .

[A1]

$\mathbb{R}$ with its usual topology is metrizable and hence Hausdorff (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y) = |x-y|$ is a metric, its open balls are the intervals $(x-r, x+r)$ , and it is unbounded, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not, Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not); $\mathbb{R}^2$ carries the product topology (The product set $\prod_{i \in I} X_i$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space, For $n \ge 1$ the product topology on $n$ copies of the usual topology of $\mathbb{R}$ is the metric topology of $d_\infty$ on $\mathbb{R}^n$ , and hence also of $d_1$ and $d_2$ , so $\mathbb{R}^n$ as a product and $\mathbb{R}^n$ as a metric space are one space).

[L1]

The function $f$ above has graph closed in $\mathbb{R}^2$ and is not continuous at $0$ (FALSE: every function between topological spaces whose graph is closed in the product is continuous).

[L2]

The reciprocal $x \mapsto 1/x$ is continuous at every $c \ne 0$ as a function on $\{\, x \in \mathbb{R} : x \ne 0 \,\}$ (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, claims 4 and 5), continuity being the $\varepsilon$ - $\delta$ condition of 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; and for a real function that condition at a point is continuity at that point as a map of topological spaces (Dictionary: for $A \subseteq \mathbb{R}$ with the metric $d(x,y) = |x-y|$ , continuity and uniform continuity of $f : A \to \mathbb{R}$ agree with the metric-space notions, the Lipschitz and Hölder conditions are the metric ones instantiated, and a subset of $\mathbb{R}$ is compact in the open-cover sense of $\mathbb{R}$ exactly when it is a compact metric subspace, claim 1, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not, Continuity of a map of topological spaces at a point and globally).

[L3]

If the codomain is compact and the graph is closed then the map is continuous (A map into a compact space whose graph is closed is continuous; so for a compact Hausdorff codomain, continuity and closedness of the graph are equivalent, claim 1, Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right).

Counterexample

technique · constructive

1.1

Take $f : \mathbb{R} \to \mathbb{R}$ with $f(x) = 1/x$ for $x \ne 0$ and $f(0) = 0$ , and give $\mathbb{R}$ its usual topology; the codomain is then Hausdorff.

A1construct

1.2

$G_f$ is closed in $\mathbb{R}^2$ and $f$ is not continuous at $0$ .

2.1

$f$ is continuous at every $c \ne 0$ : given a real $\varepsilon > 0$ , [L2] supplies a real $\delta > 0$ such that $x \ne 0$ and $|x - c| < \delta$ imply $|1/x - 1/c| < \varepsilon$ ; put $\delta' := \min\{\delta, |c|\} > 0$ , and then every $x \in \mathbb{R}$ with $|x - c| < \delta'$ satisfies $x \ne 0$ , hence $|f(x) - f(c)| = |1/x - 1/c| < \varepsilon$ . So $f$ is continuous at $c$ in the sense of 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, hence at $c$ as a map of topological spaces.

step 1.1L2

3.1

By steps 1.1, 1.2 and 2.1 the map $f$ has a closed graph and a Hausdorff codomain and is not continuous, its set of discontinuities being exactly $\{0\}$ ; so the claim is false.

step 1.1step 1.2step 2.1

4.1

By [L3] the same three facts show that $\mathbb{R}$ with its usual topology is not compact, so the hypothesis the claim should have carried is compactness of the codomain and not any separation property of it.

step 1.2step 2.1step 3.1L3discharge-construct∎

Remarks

Adding a separation hypothesis to the codomain cannot repair the claim, and this is why. In A map into a compact space whose graph is closed is continuous; so for a compact Hausdorff codomain, continuity and closedness of the graph are equivalent the Hausdorff condition is what makes a continuous map have closed graph, and compactness is what makes a closed-graph map continuous. The two hypotheses belong to opposite directions, and the witness above has the first without the second.
The failure is a single point, and it is not removable by redefining $f$ there. No value at $0$ makes $f$ continuous, because $|f(x)|$ exceeds every bound as $x$ approaches $0$ ; and no value at $0$ destroys the closedness of the graph. The example is therefore not a matter of a badly chosen value: it is the behaviour of the reciprocal near $0$ , and a compact codomain is exactly what would forbid that behaviour.
Where this sits relative to the functional-analytic closed graph theorem. That theorem replaces compactness of the codomain by completeness of both spaces and linearity of the map, and neither hypothesis is available or claimed here; the witness above is not linear, and nothing on this page bears on the functional-analytic statement.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 152 results over 21 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

Closed graph theorem (Wikipedia) (standard reference, not scraped)
Compact space (Wikipedia) (standard reference, not scraped)
Analysis 3103, Handout 7 (UCL) (standard reference, not scraped)

Refuted: a function into a Hausdorff space whose graph is closed is continuous. The function equal to 1/x1/x1/x off 000 and to 000 at 000 has a closed graph, is discontinuous at 000 alone, and has a Hausdorff codomain