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LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-10
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Hilbert cube has a bimeasurable real coding

Statement

There is an explicit Borel measurable bijection c:[0,1]NC onto a Borel subset C[0,1], whose inverse is Borel measurable. The cube carries its product topology and its Borel sigma-algebra; indices start at zero.

Facts & Assumptions

Given: The cube Q=[0,1]N with its product topology and Borel sigma-algebra; natural indices start at zero.

[F1]

The integer part is the unique integer m with mt<m+1. (Integer part: for every real x there is exactly one integer m with mx<m+1)

[F2]

Geometric series with ratios 1/2 and 1/3 converge, with their stated sums. (For r<1, k0rk=1/(1r), and for r1 the series diverges)

[F3]

The explicit map J(i,l)=2i(2l+1) bijects pairs of naturals with positive integers. (N×NN)

[F5]

Rational intervals generate the real Borel sigma-algebra. (Seven generating families for the Borel sigma-algebra on the real line)

[F6]

The rationals have an explicit countable enumeration. (Q is countably infinite)

[F7]

Between distinct reals lies a rational. (The rationals embed densely in the reals)

[F9]

Borel sets are the sigma-algebra generated by open sets. (The Borel sigma-algebra of a topological space)

[F10]

Positive-base integer powers and their reciprocals are defined. (Integer powers am)

Proof

technique · direct
1.1

For 0t1/2 put mk=2kt for k0 and bk=mk2mk1 for k1. Since mk12k1t<mk1+1, we have bk{0,1}. Each mk is Borel: {mk=l}=[l2k,(l+1)2k)[0,1/2]. Thus each digit is Borel.

F1F5F10
2.1

Here m0=0. Telescoping gives k=1nbk2k=2nmn and 0t2nmn<2n, so t=k1bk2k. The digits cannot be eventually all ones: such a tail would make t dyadic, whereas for dyadic t the integers 2kt are exact for all sufficiently large k, giving bk=0. Thus there are infinitely many zeros, including when t=0 or t=1/2.

step 1.1F2
2.2

Interleave cube digits by aJ(i,k1)=bk(xi/2). The bijection in [F3] assigns exactly one digit to each positive position. For any binary sequence a, define T(a)=j12aj3j. Its values lie in [0,1]. Agreement through position n gives T(a)T(a)3n. If the first differing position is n, its contribution has magnitude 23n and the remaining tail has magnitude at most 3n, so T(a)T(a)3n>0. Thus T is continuous and injective.

F2F3step 1.1
3.1

Conversely let (bk) be a binary sequence with infinitely many zeros and sum t1/2. Its tail after n, multiplied by 2n, lies in [0,1): the all-one tail sums to one, and at least one digit is zero. Hence 2nt=k=1nbk2nk, recovering exactly the digits of step 1.1. In binary sequence space D={0,1}N1, the allowable row set is R=N1kN{b:bk=0}{b:k1bk2k1/2}. It is Borel: cylinders are clopen and the sum is continuous because the tail is at most 2n.

step 1.1step 2.1F2F9
3.2

For a binary word w of length n, let sw=j=1n2wj3j and Iw=[sw,sw+3n]. Distinct words of the same length give disjoint intervals separated by a positive gap. The closed set K=n1w=nIw equals T[D]: each point of the intersection has a unique word at each length; nesting forces consistent prefixes; the resulting sequence has sum equal to the point since interval lengths tend to zero. Conversely each sum lies in every prefix interval. The inverse digits are continuous on K because the finitely many cylinders at each length are separated. Hence T:DK is a homeomorphism, without an appeal to product compactness.

step 2.2F2
4.1

The deinterleaving row maps DD, a(aJ(i,k1))k1, are continuous: a finite row-cylinder condition is a finite cylinder condition on a. Therefore H={a:every deinterleaved row belongs to R} is Borel in D. The homeomorphism gives C=T[H]=(T1)1[H] Borel in K. Since K is closed in [0,1], a trace Borel set in K is Borel in [0,1]: the trace sets form a sigma-algebra, and relative opens are traces of ambient opens.

step 3.1step 3.2F9
5.1

The map c(x)=T(a(x)) is measurable: every interleaved digit is Borel by step 1.1 and every finite sum has finite range with Borel level sets (finite unions of intersections of digit level sets), so [F4] applies. Its inverse on C is xi=2k1(T1z)J(i,k1)2k. These coordinates are measurable by [F4]. They lie in [0,1] and recover both compositions by the row characterization; thus c is a bijection onto precisely C.

step 1.1step 3.1step 2.2step 4.1F4
6.1

For completeness, rational intervals restricted to [0,1] form a countable basis by density. Finite coordinate boxes from these intervals are countable explicitly. Enumerate rational endpoints by [F6]. A coordinate condition xi(pj,pk)[0,1] has code J(i,J(j,k)). A list of condition codes (z0,,zl1) has code J(l,el), where e0=0 and er+1=J(er,zr). Inverting the injective J recovers the length and every entry, so this encodes lists injectively. Each box is represented by such a finite list; assigning the least code of its representations injects the family of boxes into the naturals. The empty list represents the whole cube. Every open subset of the cube is a union of a subfamily of this countable basis. Thus the Borel sigma-algebra equals the coordinate-generated sigma-algebra, and coordinate measurability in step 5.1 proves measurability of the whole inverse. This establishes all assertions.

F3F5F6F7F8F9step 5.1

Source notes

Durrett, Probability: Theory and Examples, 5th ed., Theorem 2.1.22, printed pp.53–54 (PDF pp.61–62). The complete coding paragraph and its caveat were read. The present proof replaces the abbreviated digit argument by a Borel row condition and separated ternary cylinders. The interleaving uses the actual bijection in the local supplier rather than attributing a diagonal formula to that supplier.

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