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Every at most countable subset of $R$ has measure zero

Statement

Every at most countable set $A \subseteq R$ (Finite, countably infinite, countable, uncountable) has measure zero (Measure zero (a countable cover by intervals of total length below every $ε$ ) and content zero (a finite such cover)).

The cover is explicit: the $k$ -th point of a listing of $A$ is put inside an interval of length $ε \cdot 2^{- k - 1}$ , and the lengths sum to $ε$ by For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges. No choice principle is used: a listing of $A$ is a single object, fixed once (A nonempty set is at most countable iff it is a surjective image of $N$ ), and everything after that is a formula in $k$ .

Facts & Assumptions

Given: An at most countable set $A \subseteq R$ and a real $ε > 0$ . Throughout, $θ : = 2^{- 1}$ .

[L1]

$A$ is null when for every real $ε > 0$ there are sequences $(a_{k})$ , $(b_{k})$ with $a_{k} \leq b_{k}$ , $A \subseteq ⋃_{k} [a_{k}, b_{k}]$ , and $\sum_{k < n} (b_{k} - a_{k}) \leq ε$ for every $n \in N$ (Measure zero (a countable cover by intervals of total length below every $ε$ ) and content zero (a finite such cover)).

[L2]

$[c, d] = {x : c \leq x \leq d}$ has length $d - c$ when $c \leq d$ , and $[c, c] = {c}$ has length $0$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

[L3]

A nonempty at most countable set admits a surjection $s : N \to A$ (A nonempty set is at most countable iff it is a surjective image of $N$ , Finite, countably infinite, countable, uncountable).

[L4]

Powers and the geometric series: $θ^{0} = 1$ , $θ^{k + 1} = θ^{k} θ$ , $θ^{k} > 0$ , and $\sum_{k = 0}^{\infty} θ^{k} = 2$ for $θ = 2^{- 1}$ ; a series of nonnegative terms has all its partial sums at most its sum (Integer powers $a^{m}$ , For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges, Series, partial sums, convergence and the sum, divergence, and the tail series, A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum).

[L5]

Finite sums: scaling by a constant, and $\sum_{k < n} 0 = 0$ (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L6]

Ordered-field arithmetic: $0 < 1$ , so $2 > 0$ , $4 > 0$ and $t \cdot 4^{- 1} > 0$ for $t > 0$ ; 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.

Proof

technique · direct

1.1

Let the real $ε > 0$ be given. If $A = \emptyset$ , the constant sequences $a_{k} : = 0$ and $b_{k} : = 0$ satisfy $A \subseteq ⋃_{k} [0, 0]$ vacuously and $\sum_{k < n} (b_{k} - a_{k}) = 0 \leq ε$ for every $n$ by [L5], so the condition of [L1] holds at this $ε$ . Assume from now on that $A \neq \emptyset$ and, by [L3], fix a surjection $s : N \to A$ .

givenL1L2L3L5choose

2.1

Put $δ_{k} : = ε \cdot 4^{- 1} \cdot θ^{k}$ , a positive real by [L4] and [L6], and $a_{k} : = s (k) - δ_{k}$ , $b_{k} : = s (k) + δ_{k}$ ; then $a_{k} \leq b_{k}$ and $s (k) \in [a_{k}, b_{k}]$ by [L6], so $A = {s (k) : k \in N} \subseteq ⋃_{k} [a_{k}, b_{k}]$ by step 1.1. The length of $[a_{k}, b_{k}]$ is $b_{k} - a_{k} = 2 δ_{k} = ε \cdot 2^{- 1} \cdot θ^{k}$ by [L2] and [L6].

step 1.1L2L4L6

3.1

For every $n \in N$ , $\sum_{k < n} (b_{k} - a_{k}) = ε \cdot 2^{- 1} \sum_{k < n} θ^{k} \leq ε \cdot 2^{- 1} \cdot 2 = ε$ , using scaling from [L5] and the bound on the partial sums of the geometric series from [L4].

step 2.1L4L5L6

4.1

So for every real $ε > 0$ the sequences of step 2.1 cover $A$ with all partial total lengths at most $ε$ , which by [L1] is exactly the statement that $A$ has measure zero; the empty case was settled in step 1.1.

step 1.1step 2.1step 3.1L1∎

Remarks

Indexing. Sequences here start at $k = 0$ , and the first interval has length $ε \cdot 2^{- 1}$ , not $ε$ . The total is $ε \cdot 2^{- 1} \sum_{k \geq 0} 2^{- k} = ε$ exactly, so the cover is as tight as the definition allows and nothing is wasted at the first index.
Repetitions are harmless. A surjection $s$ may repeat values, and a finite set is covered by infinitely many intervals, most of them redundant. This is why the listing form of countability (A nonempty set is at most countable iff it is a surjective image of $N$ ) is the convenient one: no injectivity and no case split between the finite and the countably infinite case is needed.
The converse fails badly. The Cantor set is uncountable and null (The Cantor set is compact, perfect, uncountable, nowhere dense and of measure zero, and it contains no interval of positive length, so its only nonempty connected subsets are single points), so "null" is very far from "countable"; and the Smith-Volterra-Cantor set is uncountable and not null (The Smith-Volterra-Cantor set is compact, perfect and nowhere dense, and does not have measure zero), so cardinality decides nothing either way.
Density decides nothing either. $Q$ is countable, hence null, and is dense in $R$ ( $Q$ is dense in $R$ and has measure zero ↗); a null set may therefore meet every interval.

Depends on

Used by

Dependency tree · two levels

55 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)

Every at most countable subset of RR has measure zero