LemmaStatement: AI-adaptedProof: AI-generatedSession-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.

Under choice, $\chi(x,X)$ and $\chi(X)$ are well-defined cardinals

Statement

Assuming choice, every $\chi(x,X)$ and the raw supremum $\chi(X)$ are well-defined cardinals.

Facts & Assumptions

Given: A space $X$ , a point $x\in X$ , and the definitions in Under choice, weight $w(X)$ , density $d(X)$ , local character $\chi(x,X)$ , and character $\chi(X)$ as raw cardinal minima and a supremum.

[L1]

Under choice every set can be well ordered and therefore has a cardinality (The well-ordering theorem, A set equinumerous with some ordinal has a least such ordinal, that ordinal is a cardinal, and equinumerous sets get the same one; no choice principle is used).

[L2]

Every nonempty set of ordinals, and hence every nonempty set of cardinals, has a least member (Trichotomy and well-ordering of the ordinals).

[L3]

Cardinals are initial ordinals, a set of ordinals has union as its least upper bound, and mutual injections give a bijection (Cardinal (initial ordinal) and cardinality, Basic closure properties of ordinals, The Schröder-Bernstein theorem).

Proof

technique · direct

1.1

The neighbourhood filter at $x$ is a local base, so local-base cardinalities form a nonempty set.

given

2.1

The candidate cardinalities are ordinals, so their nonempty set has a least member, namely $\chi(x,X)$ .

step 1.1L1L2

3.1

Let $K=\{\chi(x,X):x\in X\}$ and $\delta=\bigcup K$ . This is an ordinal and the least ordinal upper bound of $K$ by [L3]. It is a cardinal: if $\beta<\delta$ and $\beta\approx\delta$ , choose $\kappa\in K$ with $\beta<\kappa\le\delta$ . Then $\beta\preceq\kappa\preceq\delta\approx\beta$ , so [L3] gives $\beta\approx\kappa$ , contradicting that $\kappa$ is a cardinal. Thus $\delta$ is the cardinal supremum $\chi(X)$ .

L3∎

Depends on

Used by

Cited to discharge well-definedness by Under choice, weight w(X), density d(X), local character χ(x,X), and character χ(X) as raw cardinal minima and a supremum.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 64 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

D. H. Fremlin, Measure Theory, Chapter 5A (standard reference, not scraped)

Under choice, χ(x,X)\chi(x,X)χ(x,X) and χ(X)\chi(X)χ(X) are well-defined cardinals