Alphabeta Math
DefinitionDefinition: AI-adaptedProof: Not applicablejudge pass (z-ai/glm-5.2)audited 2026-07-27
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The Cantor function on [0,1], defined on the Cantor set through ternary digits and extended constantly across each removed interval

Definition

Let C be the Cantor set, D the set of sequences with values in {0,2} and Φ:D→C the bijection Φ(a)=∑k≥0ak3−k−1 of The Cantor set is exactly the set of ∑k≥1ak3−k with every ak∈{0,2}, and this gives a bijection with {0,1}N. Since Φ is a bijection it has a two-sided inverse Φ−1:C→D, and that inverse is a single function, determined and not selected (Injection, surjection, bijection).

On the Cantor set. For x∈C write a:=Φ−1(x) and put

γ(x)  :=  ∑k=0∞(ak⋅2−1) 2−k−1.

Each coefficient ak⋅2−1 is 0 or 1, so all the terms are nonnegative and every partial sum is at most ∑k<n2−k−1≤∑k=0∞2−k−1=1 (For ∣r∣<1, ∑k≥0rk=1/(1−r), and for ∣r∣≥1 the series diverges, Integer powers am, Laws of integer exponents); hence the series converges and γ(x)∈[0,1] (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, Intervals of R: the nine order-convex forms, nondegeneracy, and length). In words: γ halves each ternary digit of x and reads the result as a binary expansion.

On all of [0,1]. The Cantor function is c:[0,1]→R,

c(x)  :=  sup⁡{ γ(t):t∈C and t≤x }.

The supremum exists and is a single real number. The set on the right is nonempty, because 0∈C (The Cantor middle-thirds set as the intersection of the sets Cn obtained by removing open middle thirds) and 0≤x, and it is bounded above by 1, because γ takes values in [0,1]; so it has a least upper bound by completeness (Complete ordered field (least-upper-bound property), Lower bound, bounded below, bounded set), and that bound is unique (Suprema and infima are unique). Since 0≤γ(0)≤c(x)≤1, the values of c lie in [0,1].

That c really extends γ, that is, c(t)=γ(t) for every t∈C, is not an observation but a small theorem: it needs γ to be nondecreasing along C. It is claim 1 of The Cantor function is well defined, satisfies c(x)≤c(y) whenever x≤y, is surjective onto [0,1], and is constant on every interval removed from the Cantor set ↗, recorded in this item's justified_by, and until it is proved the two symbols are kept apart.

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