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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-26
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  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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Every open subset of Rn is the union of a countable pairwise disjoint family of dyadic cubes

Statement

Let n≥1 and let U⊆Rn be open in the metric topology of (Rn,d2) (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, Rn as the set of functions n→R, and d1, d2, d∞ are metrics on it). Then there is an at most countable family M of pairwise disjoint dyadic cubes (Dyadic cubes of generation k in Rn, Finite, countably infinite, countable, uncountable) with

U  =  ⋃M.

For U=∅ the family is empty. No choice principle is used: the cube attached to a point is the one of least generation that fits inside U, and least is a definition.

Facts & Assumptions

Given: A natural number n≥1 and an open subset U⊆Rn.

[L1]

Every x∈Rn lies in exactly one dyadic cube of generation k (For each generation, the dyadic cubes of that generation are pairwise disjoint and cover Rn).

[L2]

If Q and Q′ are dyadic cubes of generations k≤k′ and Q∩Q′≠∅, then Q′⊆Q (Two dyadic cubes are either disjoint or one contains the other).

[L3]

Qk,m={ x∈Rn:mi2−k<xi≤(mi+1)2−k for every i<n }, and every dyadic cube is nonempty (Dyadic cubes of generation k in Rn, Integer powers am).

[L4]

A nonempty box determines its parameter pair, since B determines ai and bi for every i (Half-open boxes in Rn and their volume).

[F1]

A subset U⊆X is open in (X,d) if for every x∈U there is a real r>0 with B(x,r)⊆U, where B(x,r):={ y∈X:d(x,y)<r } (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, Open ball, closed ball and sphere in a metric space).

[F2]

d2(x,y):= ∑k<n(xk−yk)2  and d∞(x,y):=max⁡{ ∣xk−yk∣:k<n } are metrics on Rn for n≥1 (Rn as the set of functions n→R, and d1, d2, d∞ are metrics on it).

[F3]

For every x∈Rn, ∥x∥2≤∥x∥1 and ∥x∥1≤n ∥x∥∞, n being the canonical natural of R; and ∥x−y∥2=d2(x,y), ∥x−y∥∞=d∞(x,y) (The finite and reverse triangle inequalities for a norm; and for n≥1 every norm N on Rn satisfies N(x)≤C∥x∥1 and is Lipschitz, hence continuous, for d2, claim 3; Each ∥⋅∥p is a norm on Rn, and the induced metrics are exactly d1, d2 and d∞ of the published metric-spaces page, claim 3; The p-norms ∥x∥p for rational p≥1, and ∥x∥∞).

[F5]

Every nonempty subset S⊆N has a least element (The well-ordering principle).

[F7]

If A and B are at most countable then so is A×B (A product of two at most countable sets is at most countable); and if A is at most countable and B⊆A then B is at most countable (Every subset of an at most countable set is at most countable, Finite, countably infinite, countable, uncountable).

Proof

technique · direct
1.1L3L4F6F7

The set D of all dyadic cubes is at most countable: each of its members is nonempty and so determines its parameter pair, whose entries mi2−k and (mi+1)2−k are rational, so the assignment of a cube to that pair is an injection of D into Qn×Qn, a countable set, and D is therefore equinumerous with an at most countable subset of it.

1.2L1L3F1F2F3F4

For every x∈U there is a natural number k such that the generation-k dyadic cube containing x is a subset of U: openness supplies a real r>0 with B(x,r)⊆U; since (2−k)k∈N is null there is k with 2−k<r/n; and every y in the generation-k cube containing x has ∣yi−xi∣<2−k in each coordinate, because xi and yi lie in one parameter interval of length 2−k, so d∞(x,y)<2−k and d2(x,y)≤n d∞(x,y)<r.

2.1step 1.2L1L2L3F5

For x∈U let k(x) be the least natural number provided by step 1.2 and let Qx be the generation-k(x) dyadic cube containing x, which is unique; then x∈Qx⊆U, and Qx is maximal among the dyadic cubes contained in U, for if Qx⊆Q′⊆U with Q′ of generation k′, then Q′ contains x and is therefore the generation-k′ cube containing x, so k′≥k(x) by minimality, and then Qx∩Q′≠∅ with k(x)≤k′ gives Q′⊆Qx and hence Q′=Qx.

3.1step 1.1step 2.1L2F7∎

Put M:={ Qx:x∈U }: its members are dyadic cubes contained in U and every x∈U lies in one of them, so ⋃M=U; two members meeting each other are nested by [L2], and each being maximal in U they are equal, so the members are pairwise disjoint; and M⊆D is at most countable by step 1.1.

Depends on

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Sources