Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 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.

Cantor and Baire sequence spaces and coordinate codings

Statement

In ZF, C=2N and N=NN are Polish under the metric d(x,x)=0 and d(x,y)=2m1 when m is the first coordinate at which x and y differ. Cantor space is compact and has no isolated points. Coordinate pairing gives homeomorphisms NNN and CCN. The map

h(x)=0x(0)10x(1)10x(2)1

is a homeomorphism of N onto D={zC:z has infinitely many 1s}, and CD is at most countable.

Facts & Assumptions

[F1]

Cantor sequence space fixes the binary cylinder topology, inherited from Baire space.

[F4]

The open-neighbourhood criterion for continuity is in Continuity of a map of topological spaces at a point and globally.

Proof

Given: The two fixed sequence spaces and their cylinder topologies. No choice principle is assumed.

1.1

Symmetry and separation of d follow from the first differing coordinate. If x,y and y,z both agree through the first n coordinates, so do x,z. Consequently d(x,z)max(d(x,y),d(y,z))d(x,y)+d(y,z), proving the metric law. For n1, Nxn={y:d(x,y)<2n}, so the metric induces precisely the cylinders. A Cauchy sequence has each coordinate eventually constant: use the Cauchy bound 2k1 for coordinate k. Define x(k) to be that unique eventual value. The Cauchy bound at 2n shows all sufficiently late terms agree with x on the first n coordinates, hence converge to x. In the binary case each value remains binary.

F1F2
1.2

Let U be an open cover of C. If no finite subfamily covers it, the root cylinder is not finitely covered. Whenever Ns is not finitely covered, at least one of Ns0,Ns1 is not finitely covered: otherwise combine their two finite covers. Recursively take the least such bit. The resulting x lies in some UU, and openness gives NxnU for some n, contradicting the construction. Hence every cover has a finite subcover, as F3 requires. Only least choices from two bits were used.

F1F3
1.3

The displayed pairing is a bijection N2N: on diagonal i+j=r its values are the consecutive integers from r(r+1)/2 to (r+1)(r+2)/21, and the diagonals partition N. Define H(x)i(j)=x(i,j). Its inverse assigns x(i,j)=zi(j), so both compositions are identities coordinate by coordinate. A finite restriction on either side constrains finitely many coordinates on the other; at each point a long enough initial cylinder fixes all those coordinates. Thus both directions are continuous by F4, in the binary case as well.

F1F4
1.4

Each block in h ends in 1 and has positive length, so h(x)D. Conversely for zD, let pn be the position of its nth 1, indexed starting at zero, obtained by successive least search. Put x(0)=p0 and x(n+1)=pn+1pn1. These are natural numbers and the block concatenation reconstructs z. Reading block lengths from h(x) returns x, giving a two-sided inverse. Fixing enough input coordinates to finish the first k output bits proves continuity of h; fixing through the kth separator proves continuity of its inverse on D.

F1F4
2.1

Finite words admit an explicit natural-number coding: encode a finite word by its length and recursively pair its entries, using i,j=(i+j)(i+j+1)/2+j. Appending infinitely many zeros gives a countable family meeting every cylinder in either space. Thus they are separable; combined with step 1.1 this proves Polishness. Given any binary cylinder containing x, change the next unrestricted bit of x and keep all other bits. The resulting different point is in the same cylinder, so no point is isolated.

F1F2step 1.1
3.1

If zD, it has only finitely many 1s. Associate the integer c(z)=j:z(j)=12j. Distinct finite binary supports give distinct sums: at their largest differing index r, the term 2r exceeds the sum j<r2j=2r1. Thus c is an injection into N, with the zero sequence mapped to zero. This proves the countability assertion and completes all constructions. QED.

step 1.4algebra

Depends on

Used by

Dependency tree · two levels

21 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