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False statementConstruction: AI-adaptedVerification: AI-adaptedprecheck passverified 2026-08-04 (gpt-5.6-sol-codex-subscription)
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FALSE: the projections of a product are closed maps

Statement

False claim: every projection πj:∏iXi→Xj of a product with the product topology is a closed map, that is, carries closed subsets of the product to closed subsets of Xj (The product set ∏i∈IXi 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, Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological).

What is true is the corresponding statement for open sets: every projection is an open map (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, claim 3). The claim above fails already for the binary product R2=R×R, whose product topology is the usual one (For n≥1 the product topology on n copies of the usual topology of R is the metric topology of d∞ on Rn, and hence also of d1 and d2, so Rn as a product and Rn as a metric space are one space). The witness is the hyperbola

H  :=  { (x,y)∈R2:xy=1 },

which is closed in R2 while π0[H]=R∖{0} is not closed in R.

Facts & Assumptions

Given: R2=∏k<2R with the product topology, the first projection π0(x,y)=x, and the set H of the statement.

[A1]

The product topology on R2 is the metric topology of d∞((x,y),(x′,y′))=max⁡{∣x−x′∣, ∣y−y′∣}, so R2 is metrizable (For n≥1 the product topology on n copies of the usual topology of R is the metric topology of d∞ on Rn, and hence also of d1 and d2, so Rn as a product and Rn as a metric space are one space).

[L1]

The multiplication map m:R2→R, m(x,y):=xy, is continuous. Indeed, at (a,b) and for ε>0, put δ:=min⁡{1, ε∣a∣+∣b∣+1}>0. If d∞((x,y),(a,b))<δ, then ∣x∣<∣a∣+1 and ∣xy−ab∣≤∣x∣ ∣y−b∣+∣b∣ ∣x−a∣<(∣a∣+∣b∣+1)δ≤ε. The bound uses xy−ab=x(y−b)+b(x−a), the triangle inequality ∣u+v∣≤∣u∣+∣v∣ (The triangle inequality) and ∣uv∣=∣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: if t≠1, the open interval of radius ∣t−1∣/2 about t 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⊆R is open in the usual topology exactly when every point of U has a bounded open interval around it inside U; (a,b)={t:a<t<b} (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).

Refutation

technique · direct
1.1

Since H=m−1[{1}], [L1] and [L2] show that H is closed in R2.

A1L1L2
1.2

For x≠0 the point (x,1/x) lies in H, since x⋅(1/x)=1; and (0,y)∉H for every y, since 0⋅y=0≠1. So π0[H]=R∖{0}.

given
1.3

R∖{0} is not closed in R: its complement {0} is not open, because for every r>0 the interval (−r,r) contains r/2, which is nonzero and hence outside {0}.

L3
2.1

By step 1.1 the set H is closed in R2, while by steps 1.2 and 1.3 its image π0[H]=R∖{0} is not closed in R; so π0 is not a closed map by [A2] and the claim is false.

step 1.1step 1.2step 1.3A2∎

Remarks

Depends on

Used by

Dependency tree · two levels

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Sources