Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27
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Q\mathbb{Q} is covered by open intervals of total length ε\varepsilon, for every ε>0\varepsilon > 0

Example

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

QR⊆⋃k∈NVkand∑k=0∞length⁡(Vk)=ε.\mathbb{Q}_{\mathbb{R}} \subseteq \bigcup_{k \in \mathbb{N}} V_k \qquad \text{and} \qquad \sum_{k=0}^{\infty} \operatorname{length}(V_k) = \varepsilon .

Explicitly, if e:N→QRe : \mathbb{N} \to \mathbb{Q}_{\mathbb{R}} is a bijection, one may take Vk:=(e(k)−ε2−k−2, e(k)+ε2−k−2)V_k := \big(e(k) - \varepsilon 2^{-k-2},\ e(k) + \varepsilon 2^{-k-2}\big), of length ε2−k−1\varepsilon 2^{-k-1}.

This is Every at most countable subset of R\mathbb{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\mathbb{R} is nonetheless covered by open intervals whose lengths add up to a millionth.

Facts & Assumptions

Given: A real ε>0\varepsilon > 0 and the set QR\mathbb{Q}_{\mathbb{R}} of rationals inside R\mathbb{R}.

[L1]

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

[L2]

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

[L3]

∑k=0∞2−k=2\sum_{k=0}^{\infty} 2^{-k} = 2, powers satisfy 2−k−1=2−12−k2^{-k-1} = 2^{-1}2^{-k}, and a convergent series of nonnegative terms has sum the supremum of its partial sums (For ∣r∣<1|r| < 1, ∑k≥0rk=1/(1−r)\sum_{k \ge 0} r^k = 1/(1-r), and for ∣r∣≥1|r| \ge 1 the series diverges, Integer powers ama^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\mathbb{R} has measure zero, and nullity means a cover by closed intervals whose partial total lengths stay below ε\varepsilon (Every at most countable subset of R\mathbb{R} has measure zero, Measure zero (a countable cover by intervals of total length below every ε\varepsilon) and content zero (a finite such cover)).

[L6]

Ordered-field arithmetic: 0<10 < 1, so 2>02 > 0, 4>04 > 0 and ε2−k−2>0\varepsilon 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→QRe : \mathbb{N} \to \mathbb{Q}_{\mathbb{R}} and put δk:=ε⋅2−k−2\delta_k := \varepsilon \cdot 2^{-k-2}, a positive real by [L3] and [L6], and Vk:=(e(k)−δk, e(k)+δk)V_k := (e(k) - \delta_k,\ e(k) + \delta_k), an open interval of length 2δk=ε2−k−12\delta_k = \varepsilon 2^{-k-1} by [L2], [L3] and [L6].

givenL1L2L3L6choose
1.2

The family covers QR\mathbb{Q}_{\mathbb{R}}: every rational is e(k)e(k) for some kk, and e(k)∈Vke(k) \in V_k because e(k)−δk<e(k)<e(k)+δke(k) - \delta_k < e(k) < e(k) + \delta_k by [L6].

L1L2L6
2.1

The lengths sum to ε\varepsilon: by [L4] the partial sums are ∑k<nε2−k−1=ε2−1∑k<n2−k\sum_{k<n}\varepsilon 2^{-k-1} = \varepsilon 2^{-1}\sum_{k<n}2^{-k}, and by [L3] the series ∑k2−k\sum_k 2^{-k} converges to 22, so by [L3] and [L4] the series ∑kε2−k−1\sum_k \varepsilon 2^{-k-1} converges with sum ε⋅2−1⋅2=ε\varepsilon \cdot 2^{-1} \cdot 2 = \varepsilon.

step 1.1L3L4L6
3.1

So the open intervals VkV_k cover QR\mathbb{Q}_{\mathbb{R}} with total length exactly ε\varepsilon, as claimed; since ε>0\varepsilon > 0 was arbitrary, this also re-exhibits the nullity of QR\mathbb{Q}_{\mathbb{R}} given by [L5], the closed intervals [e(k)−δk,e(k)+δk][e(k) - \delta_k, e(k) + \delta_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 VkV_k is open, so ⋃kVk\bigcup_k V_k is an open set containing every rational, hence dense; and its covering intervals have total length ε\varepsilon. Iterating this over a sequence of shrinking ε\varepsilon is exactly the construction of R\mathbb{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⋅2−1\varepsilon \cdot 2^{-1}, not ε\varepsilon: sequences here start at k=0k = 0 and the total ε2−1∑k≥02−k\varepsilon 2^{-1} \sum_{k \ge 0} 2^{-k} is exactly ε\varepsilon. Copying the classical ε2−k\varepsilon 2^{-k} from a 11-indexed source would give total 2ε2\varepsilon.

  • What this does not show. It does not show that the union of the VkV_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 VkV_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\varepsilon) and content zero (a finite such cover) is phrased in terms of covers throughout.

Depends on

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