CounterexampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)verified 2026-08-04 (gpt-5.6-sol-codex-subscription)

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.

The hyperbola ${(x, y) : x y = 1}$ is closed in $R^{2}$ and its image under the first projection is $R ∖ {0}$ , which is not closed

Statement refuted

Refuted: that the projections of a product with the product topology are closed maps (FALSE: the projections of a product are closed maps, Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological).

Witness. In $R^{2} = R \times R$ with the product topology, which is the usual topology (For $n \geq 1$ the product topology on $n$ copies of the usual topology of $R$ is the metric topology of $d_{\infty}$ on $R^{n}$ , and hence also of $d_{1}$ and $d_{2}$ , so $R^{n}$ as a product and $R^{n}$ as a metric space are one space, $R^{n}$ as the product of $n$ copies of the real line: the product topology is the Euclidean topology and the projections are continuous, open and surjective), take

$H : = {(x, y) \in R^{2} : x y = 1} .$

Then $H$ is closed in $R^{2}$ , its image under the first projection is $π_{0} [H] = R ∖ {0}$ , and $R ∖ {0}$ is not closed in $R$ : the point $0$ lies in its closure and not in it. So $π_{0}$ is not a closed map, although it is a continuous open surjection (A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice).

Facts & Assumptions

Given: $R^{2}$ with the product topology, the first projection $π_{0} (x, y) = x$ , and the set $H$ above.

[A1]

The product topology on $R^{2}$ is the metric topology of $d_{\infty} ((x, y), (x^{'}, y^{'})) = \max {∣ x - x^{'} ∣, ∣ y - y^{'} ∣}$ , and $R^{2}$ is therefore metrizable (For $n \geq 1$ the product topology on $n$ copies of the usual topology of $R$ is the metric topology of $d_{\infty}$ on $R^{n}$ , and hence also of $d_{1}$ and $d_{2}$ , so $R^{n}$ as a product and $R^{n}$ as a metric space are one space).

[A2]

$f$ is a closed map when images of closed sets are closed; every projection is continuous and open (Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological, A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice).

[L1]

The multiplication map $m : R^{2} \to R$ , $m (x, y) : = x y$ , is continuous. At $(a, b)$ and for $ε > 0$ , take $δ : = \min {1, \frac{ε}{∣ a ∣ + ∣ b ∣ + 1}} > 0.$ If $d_{\infty} ((x, y), (a, b)) < δ$ , then $∣ x ∣ < ∣ a ∣ + 1$ and $∣ x y - a b ∣ \leq ∣ x ∣ ∣ y - b ∣ + ∣ b ∣ ∣ x - a ∣ < (∣ a ∣ + ∣ b ∣ + 1) δ \leq ε .$ The bound uses $x y - a b = x (y - b) + b (x - a)$ , the triangle inequality $∣ u + v ∣ \leq ∣ u ∣ + ∣ v ∣$ (The triangle inequality) and $∣ u v ∣ = ∣ u ∣ ∣ v ∣$ (Basic properties of the absolute value). This is the metric definition of continuity (Continuity of a map between metric spaces, at a point and globally, in the $ε$ - $δ$ form, Inverses of positives are positive, and reciprocation reverses order, Maximum and minimum of a set).

[L2]

The singleton ${1}$ is closed in $R$ : the open interval of radius $∣ t - 1 ∣ / 2$ about any $t \neq 1$ avoids $1$ . A continuous map of metric spaces has closed preimages of closed sets (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, Metric continuity characterisations, with countable choice for the sequential converse, clause (c)).

[L3]

$U \subseteq R$ is open exactly when every point of $U$ has a bounded open interval around it inside $U$ (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).

Counterexample

technique · direct

1.1

Since $H = m^{- 1} [{1}]$ , [L1] and [L2] show that $H$ is closed in $R^{2}$ .

A1L1L2

1.2

For $x \neq 0$ one has $(x, 1 / x) \in H$ , since $x \cdot (1 / x) = 1$ ; and no point $(0, y)$ lies in $H$ , since $0 \cdot y = 0 \neq 1$ . Hence $π_{0} [H] = R ∖ {0}$ .

given

1.3

The set $R ∖ {0}$ is not closed: its complement ${0}$ is not open, because every interval $(- r, r)$ with $r > 0$ contains the nonzero point $r / 2$ .

2.1

By step 1.1 the set $H$ is closed and by steps 1.2 and 1.3 its image $π_{0} [H]$ is not, so $π_{0}$ is not a closed map by [A2]; by [A2] it is nevertheless a continuous open map, which refutes the claim.

step 1.1step 1.2step 1.3A2∎

Remarks

The picture behind the computation. The two branches of $H$ run away to infinity as $x$ approaches $0$ , so the first coordinates of points of $H$ come arbitrarily close to $0$ while $H$ itself stays away from the whole vertical axis: the point of $H$ nearest to $(0, y_{0})$ with $y_{0} \neq 0$ is at horizontal distance about $1 / ∣ y_{0} ∣$ , and $H$ approaches the axis only at unbounded heights. Projecting forgets the height, so the first coordinates alone fill $R ∖ {0}$ and their limit point $0$ is missing from the image; nothing about $H$ being closed prevents that, because the points of $H$ whose images converge to $0$ escape to infinity instead of converging in $R^{2}$ .
Which hypothesis would repair it, and why it is not available. If the second factor were compact, the projection along it would be a closed map, and $H$ would then have to meet the axis. Compactness is later in the reading order, so no repair is stated here; what is recorded is only the failure.
The other two conclusions about projections survive untouched. $π_{0}$ is open and surjective (A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice, $R^{n}$ as the product of $n$ copies of the real line: the product topology is the Euclidean topology and the projections are continuous, open and surjective); those hold for every product and are not affected by this witness. Openness and closedness are independent properties of a map, as Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological records.

Depends on

Used by

On A = ([0,∞) × ℝ) ∪ (ℝ × {0}) the first projection is a quotient map, by the section x ↦ (x,0), and is neither open nor closed Counterexample

Dependency tree · two levels

56 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

Open and closed maps (Wikipedia) (standard reference, not scraped)
Hyperbola (Wikipedia) (standard reference, not scraped)
Product topology (Wikipedia) (standard reference, not scraped)
A quotient map which is neither open nor closed (UC Riverside Math 205A notes) (standard reference, not scraped)

The hyperbola {(x,y):xy=1}{(x,y):xy=1} is closed in R2R2 and its image under the first projection is R∖{0}R∖{0}, which is not closed