Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-31 rests on unproved material (inherited)
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.

Rests on 3 statements not proved in this library, by way of the results it cites. This item cites no such statement directly; it depends on results that do. The unproved premises it inherits are Cohen's first model: an infinite Dedekind-finite set of reals, Sierpiński 1947: the generalised continuum hypothesis implies the Axiom of Choice and The continuum hypothesis and its generalisation are independent of ZFC. Each is recorded with a citation to the literature and is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

Jones's bound: under choice, a closed discrete subspace of a normal space cannot have more subsets than a dense set has subsets

Statement

Assume the Axiom of Choice. If DD is a closed discrete subspace of a normal space XX and EXE\subseteq X is dense, then there is an injection P(D)P(E)\mathcal P(D)\to\mathcal P(E). In cardinal notation, 2D2E2^{|D|}\le 2^{|E|}.

Facts & Assumptions

Given: A normal space XX, a closed discrete DXD\subseteq X, and a dense EXE\subseteq X.

[A1]

The Axiom of Choice supplies a choice function for every family of nonempty sets (The Axiom of Choice).

[F1]

Every subset of a discrete subspace is closed in that subspace; because DD is closed in XX, each subset of DD is closed in XX (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace).

Proof

technique · direct
1.1

For every ADA\subseteq D, the sets AA and DAD\setminus A are disjoint closed subsets of XX. By normality there is an open UAU_A containing AA and an open VAV_A containing DAD\setminus A with UAVA=U_A\cap V_A=\varnothing.

F1F2
2.1

Apply [A1] to choose one such pair (UA,VA)(U_A,V_A) for every ADA\subseteq D, and define Φ(A)=UAEE\Phi(A)=U_A\cap E\subseteq E.

A1step 1.1
3.1

If ABA\ne B, take dABd\in A\setminus B after interchanging them if necessary. Then dUAVBd\in U_A\cap V_B, a nonempty open set meeting EE; a point of EUAVBE\cap U_A\cap V_B lies in Φ(A)\Phi(A) and not in Φ(B)\Phi(B).

F2step 2.1
4.1

Thus Φ\Phi is injective. By [F3], this is the asserted cardinal inequality.

F3step 3.1

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 76 results over 22 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