ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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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.

$Q$ is covered by open intervals of total length $ε$ , for every $ε > 0$

Example

Let $Q_{R} \subseteq R$ be the set of rationals (The rationals embed densely in the reals) and let $ε > 0$ be real. Then there is a sequence $(V_{k})_{k \in N}$ of open intervals (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length) with

$Q_{R} \subseteq ⋃_{k \in N} V_{k} and \sum_{k = 0}^{\infty} length (V_{k}) = ε .$

Explicitly, if $e : N \to Q_{R}$ is a bijection, one may take $V_{k} : = (e (k) - ε 2^{- k - 2}, e (k) + ε 2^{- k - 2})$ , of length $ε 2^{- k - 1}$ .

This is Every at most countable subset of $R$ has measure zero made concrete for the most familiar countable set, and it is the computation that makes measure zero look paradoxical: a set that meets every interval of $R$ is nonetheless covered by open intervals whose lengths add up to a millionth.

Facts & Assumptions

Given: A real $ε > 0$ and the set $Q_{R}$ of rationals inside $R$ .

[L1]

$Q \approx N$ and $q \mapsto \hat{q}$ is injective with image $Q_{R}$ , so there is a bijection $e : N \to Q_{R}$ ( $Q$ is countably infinite, The rationals embed densely in the reals, Equinumerous sets, $A \approx B$ and $A ⪯ B$ , Finite, countably infinite, countable, uncountable, A nonempty set is at most countable iff it is a surjective image of $N$ ).

[L2]

$(c, d)$ is an open interval of length $d - c$ , and it is an open set (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, Open subset of $R$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen).

[L3]

$\sum_{k = 0}^{\infty} 2^{- k} = 2$ , powers satisfy $2^{- k - 1} = 2^{- 1} 2^{- k}$ , and a convergent series of nonnegative terms has sum the supremum of its partial sums (For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges, Integer powers $a^{m}$ , Laws of integer exponents, A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum, Series, partial sums, convergence and the sum, divergence, and the tail series).

[L4]

Finite sums scale by a constant (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L5]

Every at most countable subset of $R$ has measure zero, and nullity means a cover by closed intervals whose partial total lengths stay below $ε$ (Every at most countable subset of $R$ has measure zero, Measure zero (a countable cover by intervals of total length below every $ε$ ) and content zero (a finite such cover)).

[L6]

Ordered-field arithmetic: $0 < 1$ , so $2 > 0$ , $4 > 0$ and $ε 2^{- k - 2} > 0$ ; adding a constant and multiplying by a positive preserve an inequality (The multiplicative identity is positive, Order is preserved by adding a constant and by adding inequalities, Sign rules for products and monotonicity of multiplication, Ordered field, Complete ordered field (least-upper-bound property)). These order-arithmetic facts are stated by their sources for the strict order only; the nonstrict forms used below follow by adjoining the equality case, in which the two sides coincide.

Verification

technique · direct

1.1

By [L1] fix a bijection $e : N \to Q_{R}$ and put $δ_{k} : = ε \cdot 2^{- k - 2}$ , a positive real by [L3] and [L6], and $V_{k} : = (e (k) - δ_{k}, e (k) + δ_{k})$ , an open interval of length $2 δ_{k} = ε 2^{- k - 1}$ by [L2], [L3] and [L6].

givenL1L2L3L6choose

1.2

The family covers $Q_{R}$ : every rational is $e (k)$ for some $k$ , and $e (k) \in V_{k}$ because $e (k) - δ_{k} < e (k) < e (k) + δ_{k}$ by [L6].

L1L2L6

2.1

The lengths sum to $ε$ : by [L4] the partial sums are $\sum_{k < n} ε 2^{- k - 1} = ε 2^{- 1} \sum_{k < n} 2^{- k}$ , and by [L3] the series $\sum_{k} 2^{- k}$ converges to $2$ , so by [L3] and [L4] the series $\sum_{k} ε 2^{- k - 1}$ converges with sum $ε \cdot 2^{- 1} \cdot 2 = ε$ .

step 1.1L3L4L6

3.1

So the open intervals $V_{k}$ cover $Q_{R}$ with total length exactly $ε$ , as claimed; since $ε > 0$ was arbitrary, this also re-exhibits the nullity of $Q_{R}$ given by [L5], the closed intervals $[e (k) - δ_{k}, e (k) + δ_{k}]$ having the same lengths.

step 1.1step 1.2step 2.1L5∎

Remarks

The union is a dense open set of arbitrarily small total length. Each $V_{k}$ is open, so $⋃_{k} V_{k}$ is an open set containing every rational, hence dense; and its covering intervals have total length $ε$ . Iterating this over a sequence of shrinking $ε$ is exactly the construction of $R$ is the union of a meager set and a set of measure zero, so smallness of category and smallness of measure are independent notions, where the intersection of countably many such open sets turns out to be null and residual at the same time.
Indexing. The first interval has length $ε \cdot 2^{- 1}$ , not $ε$ : sequences here start at $k = 0$ and the total $ε 2^{- 1} \sum_{k \geq 0} 2^{- k}$ is exactly $ε$ . Copying the classical $ε 2^{- k}$ from a $1$ -indexed source would give total $2 ε$ .
What this does not show. It does not show that the union of the $V_{k}$ is small: that union is an open set containing a dense set, and one may not conclude anything about its own total length from the lengths of the $V_{k}$ , since they overlap heavily. The correct statement is about the cover, not the union, and that is why Measure zero (a countable cover by intervals of total length below every $ε$ ) and content zero (a finite such cover) is phrased in terms of covers throughout.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

77 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

Null set (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 11 (standard reference, not scraped)
MIT 18.125, Homework 2: Measure-zero sets (standard reference, not scraped)

QQ is covered by open intervals of total length εε, for every ε>0ε>0