False statementConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)verified 2026-07-29 (claude-fable-5)

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FALSE: every function between topological spaces whose graph is closed in the product is continuous

Statement

False claim: if $X$ and $Y$ are topological spaces and $f : X \to Y$ is a function whose graph $G_{f}$ is 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).

The refutation is the function $f : R \to R$ , with $R$ carrying its usual topology (The absolute value makes $R$ 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), given by

$f (x) : = \frac{1}{x} (x \neq 0), f (0) : = 0.$

Its graph is closed in $R \times R$ and it is not continuous at $0$ . Since a map into a compact space with closed graph 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), the same witness shows as a by-product that $R$ with its usual topology is not compact (Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right): the compactness hypothesis in that theorem is what the claim above drops, and it is not redundant.

Facts & Assumptions

Given: $R$ with its usual topology, the product $R \times R$ with the product topology, the function $f$ above, and its graph $G_{f} = {z \in R \times R : z_{1} = f (z_{0})}$ .

[A1]

A set $U \subseteq R$ is open in the usual topology exactly when for every $x \in U$ there is a real $r > 0$ with $(x - r, x + r) \subseteq U$ ; in particular every bounded open interval is open (The absolute value makes $R$ 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, claims 2 and 3, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

[A2]

The boxes $U \times W$ with $U$ and $W$ open in $R$ form a basis for the product topology on $R \times R$ , the index set being $2$ (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, Basis and subbasis for a topology, and the topology generated by a family of sets).

[A3]

$f$ is continuous at $x_{0}$ exactly when for every open $V$ with $f (x_{0}) \in V$ there is an open $U ∋ x_{0}$ with $f [U] \subseteq V$ , and continuous when this holds everywhere (Continuity of a map of topological spaces at a point and globally).

[L1]

A point lies in $\overline{A}$ exactly when every basic open set containing it meets $A$ , and $A$ is closed exactly when $A = \overline{A}$ (A point lies in the closure of $A$ iff every basic neighbourhood of it meets $A$ ; the closure is the smallest closed superset and equals $A$ together with its derived set, claims 1(d) and 2).

[L2]

The reciprocal $x \mapsto 1 / x$ is continuous at every $c \neq 0$ as a function on ${x \in R : x \neq 0}$ , being the quotient of the constant function $1$ by the identity (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 at $c$ means that for every real $ε > 0$ there is a real $δ > 0$ such that $x \neq 0$ and $∣ x - c ∣ < δ$ imply $∣ 1 / x - 1 / c ∣ < ε$ (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).

[L3]

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

[L4]

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).

Refutation

technique · constructive

1.1

Define $f : R \to R$ by $f (x) : = 1 / x$ for $x \neq 0$ and $f (0) : = 0$ ; this is a function on all of $R$ , every nonzero real having a multiplicative inverse.

construct

1.2

Let $z = (a, b) \in R \times R$ with $z \notin G_{f}$ , so that $b \neq f (a)$ .

given

1.3

$f$ is not continuous at $0$ : put $V : = (- 1, 1)$ , an open set containing $f (0) = 0$ , and let $U$ be any open set with $0 \in U$ ; by [A1] there is a real $r > 0$ with $(- r, r) \subseteq U$ , by [L3] there is a natural $n \geq 1$ with $1 / n < \min {r, 1}$ , and then $1 / n \in U$ while $f (1 / n) = n > 1$ , so $f (1 / n) \notin V$ and $f [U] \subseteq̸ V$ .

A1A3L3

2.1

Suppose $a \neq 0$ , so $f (a) = 1 / a$ and $ε : = ∣ b - 1 / a ∣ / 2 > 0$ ; by [L2] fix a real $δ_{1} > 0$ such that $x \neq 0$ and $∣ x - a ∣ < δ_{1}$ imply $∣ 1 / x - 1 / a ∣ < ε$ , and put $δ : = \min {δ_{1}, ∣ a ∣} > 0$ .

step 1.2L2choose

2.2

Suppose $a = 0$ , so $f (a) = 0$ and $b \neq 0$ , hence $∣ b ∣ > 0$ ; put $ρ : = 1 / (2 ∣ b ∣) > 0$ and $σ : = ∣ b ∣ / 2 > 0$ and $B : = (- ρ, ρ) \times (b - σ, b + σ)$ , a basic open set containing $z = (0, b)$ .

step 1.2A1A2construct

3.1

With $a \neq 0$ the box $B : = (a - δ, a + δ) \times (b - ε, b + ε)$ is a basic open set containing $z$ and $B \cap G_{f} = \emptyset$ : for $(x, y) \in B$ one has $∣ x - a ∣ < δ \leq ∣ a ∣$ , so $x \neq 0$ and $f (x) = 1 / x$ with $∣ 1 / x - 1 / a ∣ < ε$ , whence $∣ f (x) - b ∣ \geq ∣ b - 1 / a ∣ - ∣ 1 / x - 1 / a ∣ > 2 ε - ε = ε > ∣ y - b ∣$ and therefore $y \neq f (x)$ .

step 2.1A1A2

3.2

With $a = 0$ the box $B$ of step 2.2 satisfies $B \cap G_{f} = \emptyset$ : let $(x, y) \in B$ with $y = f (x)$ ; if $x = 0$ then $y = 0$ and $∣ y - b ∣ = ∣ b ∣ > σ$ , contradicting $y \in (b - σ, b + σ)$ ; and if $x \neq 0$ then $0 < ∣ x ∣ < ρ$ gives $∣ f (x) ∣ = 1 / ∣ x ∣ > 1 / ρ = 2 ∣ b ∣$ , while $∣ y ∣ \leq ∣ b ∣ + ∣ y - b ∣ < ∣ b ∣ + σ = 3 ∣ b ∣ / 2 < 2 ∣ b ∣$ , contradicting $y = f (x)$ .

step 2.2

4.1

Every $z \notin G_{f}$ has a basic open set containing it and missing $G_{f}$ , by step 3.1 if its first coordinate is nonzero and by step 3.2 if it is zero; so no such $z$ lies in $\overline{G_{f}}$ , whence $\overline{G_{f}} = G_{f}$ and $G_{f}$ is closed in $R \times R$ .

step 1.2step 3.1step 3.2L1

5.1

By step 1.3 and [A3] the function $f$ is not continuous, while by step 4.1 its graph is closed; so the claim is false.

step 4.1step 1.3A3

6.1

By [L4] a function into a compact codomain with closed graph is continuous, so steps 4.1 and 1.3 also show that $R$ with its usual topology is not compact; the witness therefore refutes the claim and locates the missing hypothesis at the same time.

step 4.1step 1.3step 5.1L4discharge-construct∎

Remarks

Which hypothesis was dropped. The true statements in this neighbourhood are the two halves 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: a closed graph gives continuity when the codomain is compact, and continuity gives a closed graph when the codomain is Hausdorff (The graph of a continuous map into a Hausdorff space is closed in the product). The claim above asks for the first conclusion with neither hypothesis, and the witness has a Hausdorff codomain, so it is compactness and not separation that is missing.
Where the closedness of the graph comes from, informally. Off the vertical axis the graph is closed because the reciprocal is continuous there; on the axis it is closed because the function escapes: near $0$ the values are large in absolute value, so a small box around a point $(0, b)$ with $b \neq 0$ cannot meet the graph at all. That escape is exactly what a compact codomain would forbid.
The value chosen at $0$ is immaterial. Replacing $f (0) = 0$ by any fixed real $c$ leaves both conclusions standing. For the graph, a point $(0, b)$ with $b \neq c$ is separated from it by the box $(- ρ, ρ) \times (b - σ, b + σ)$ with $σ : = ∣ b - c ∣ / 2$ and $ρ : = 1 / (∣ b ∣ + σ)$ : the value at $0$ is $c$ , which lies outside the second factor, and for $x \neq 0$ in the first factor $∣ 1 / x ∣ > ∣ b ∣ + σ$ forces $∣ 1 / x - b ∣ > σ$ . For the discontinuity, step 1.3 uses only that $f (1 / n) = n$ exceeds every bound, which does not involve $f (0)$ at all. The value $0$ is chosen above only because it makes the two computations shortest.

Depends on

Used by

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 Counterexample

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67 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

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