Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedaudited 2026-09-10
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.

Continuous injections of sequence spaces into the real line

Statement

In ZF the map

j:2N→R,j(b)=∑k=0∞2b(k)3−k−1

is a continuous injection. If h:NN→2N is the block-coding map, then e=j∘h is a continuous injection of Baire space into R.

Facts & Assumptions

[F2]

Cantor and Baire sequence spaces and coordinate codings supplies the continuous injective block coding and the cylinder topologies.

[F3]

Continuity of a map of topological spaces at a point and globally gives the open-neighbourhood criterion for continuity.

Proof

Given: The two explicit series and block maps, with no choice assumption.

1.1F1F2

Each digit 2b(k) is zero or two, so F1 applies to give a convergent series and injectivity of j. If b,c agree in their first n coordinates, subtraction of their convergent series and the geometric tail bound give

∣j(b)−j(c)∣≤∑k=n∞2 3−k−1=3−n.

The equality follows from the finite geometric sum and its limit. Since 3n≥n+1, these tails tend to zero. Given an open neighbourhood O of j(b), take a radius ϵ>0 ball contained in it and n with 3−n<ϵ. The cylinder Nb↾n then maps into O by the inequality. Thus j is continuous by F3 and F2. [F1, F2, F3]

2.1F2F3step 1.1

By F2 h is injective and continuous. If e(x)=e(y), injectivity of j gives h(x)=h(y) and injectivity of h gives x=y. For a real open O, e−1[O]=h−1[j−1[O]] is open by continuity of both maps, proving continuity of e. In particular j sends the zero sequence to zero and the all-one sequence to one, as the same geometric sum shows. QED.

Depends on

Used by

Dependency tree · two levels

31 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