Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-31
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.

Arens space S2 is sequential but not Fréchet–Urysohn

Example

Let S2={∞}∪{xn:n∈N}∪{xn,m:n,m∈N}. The xn,m are isolated; neighbourhoods of xn contain a tail of its row; a neighbourhood of ∞ contains neighbourhoods of all but finitely many xn. Then S2 is sequential, but is not Fréchet–Urysohn.

Facts & Assumptions

Given: The displayed topology on S2 and A={xn,m:n,m∈N}.

[A1]

Fréchet–Urysohn and sequential spaces have the closure and sequential-closed meanings in Fréchet–Urysohn spaces and sequential spaces.

[L1]

Closure is characterized by meeting every neighbourhood, and convergence by eventual membership in every neighbourhood (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, Convergence and cluster points of a sequence in a topological space, sequential continuity, and the sequential closure).

[L2]

Finite subsets of N have maxima, nonempty subsets have least members, and recursion produces sequences from uniquely specified successive terms (Every nonempty finite set of reals has a maximum and a minimum, The well-ordering principle, The recursion theorem).

Verification

technique · constructive
1.1

Every neighbourhood of ∞ meets A, so ∞∈A‾ by [L1]. No sequence in A converges to ∞: if it visits a row infinitely often, a neighbourhood omitting that row defeats convergence. If it visits every row finitely, use [L2] to put the threshold on each visited row one above the maximum selected second coordinate, and threshold 0 on every unvisited row. The resulting neighbourhood omits the whole sequence.

L1L2construct
1.2

Let C be sequentially closed. An isolated closure point lies in C. If xn∈C‾, then C meets its n-th row arbitrarily far out; recursion and least elements from [L2] give a sequence of row points in C converging to xn, so xn∈C. If ∞∈C‾, then infinitely many xn lie in C: otherwise omit the finitely many rows whose centres lie in C. In every remaining row, the preceding conclusion shows that C has only finitely many points; using their maximum as in [L2] gives a canonical tail disjoint from C. These tails form a neighbourhood of ∞ disjoint from C, contradicting ∞∈C‾.

A1L1L2
2.1

Hence S2 is not Fréchet–Urysohn.

step 1.1A1
2.2

The indices n with xn∈C form an infinite subset of N; list them increasingly using [L2]. The resulting sequence of row centres converges to ∞, so sequential closedness puts ∞ in C. Thus every sequentially closed C contains all its closure points and is closed. Therefore S2 is sequential.

step 1.2A1L2
3.1

The two conclusions prove the example.

step 2.1step 2.2discharge-construct∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

30 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