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

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The Cantor function takes the value $1 / 2$ on all of $[1 / 3, 2 / 3]$ , and its values at $1 / 9$ , $1 / 4$ and $7 / 9$

Example

Let $c$ be the Cantor function (The Cantor function on $[0, 1]$ , defined on the Cantor set through ternary digits and extended constantly across each removed interval). Then

$c (x) = \frac{1}{2} for every x \in [\frac{1}{3}, \frac{2}{3}], c (\frac{1}{9}) = \frac{1}{4}, c (\frac{1}{4}) = \frac{1}{3}, c (\frac{7}{9}) = \frac{3}{4} .$

Each value is computed by halving the ternary digits and reading the result in base two, which is what The Cantor function on $[0, 1]$ , defined on the Cantor set through ternary digits and extended constantly across each removed interval prescribes on $C$ , and by the constancy across gaps of The Cantor function is well defined, satisfies $c (x) \leq c (y)$ whenever $x \leq y$ , is surjective onto $[0, 1]$ , and is constant on every interval removed from the Cantor set off $C$ .

Facts & Assumptions

Given: The Cantor set $C$ , the bijection $Φ$ of The Cantor set is exactly the set of $\sum_{k \geq 1} a_{k} 3^{- k}$ with every $a_{k} \in {0, 2}$ , and this gives a bijection with ${0, 1}^{N}$ , and the functions $γ$ and $c$ of The Cantor function on $[0, 1]$ , defined on the Cantor set through ternary digits and extended constantly across each removed interval. Write $σ a$ for the shifted sequence.

[L1]

$Φ (a) = \sum_{k \geq 0} a_{k} 3^{- k - 1}$ is a bijection from the ${0, 2}$ -valued sequences onto $C$ , and $γ (Φ (a)) = \sum_{k \geq 0} (a_{k} 2^{- 1}) 2^{- k - 1}$ (The Cantor set is exactly the set of $\sum_{k \geq 1} a_{k} 3^{- k}$ with every $a_{k} \in {0, 2}$ , and this gives a bijection with ${0, 1}^{N}$ , The Cantor function on $[0, 1]$ , defined on the Cantor set through ternary digits and extended constantly across each removed interval, Series, partial sums, convergence and the sum, divergence, and the tail series).

[L2]

Geometric tails: $\sum_{k \geq m} 2 \cdot 3^{- k - 1} = 3^{- m}$ and $\sum_{k \geq m} 2^{- k - 1} = 2^{- m}$ ; convergent series add and scale termwise, and $\sum_{k \geq 0} t_{k} = t_{0} + \sum_{k \geq 1} t_{k}$ (For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges, Convergent series add and scale termwise, 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, Integer powers $a^{m}$ , Laws of integer exponents).

[L3]

Digit sequences: $\frac{1}{3} = Φ (0, 2, 2, 2, \dots)$ , $\frac{2}{3} = Φ (2, 0, 0, 0, \dots)$ , $\frac{1}{9} = Φ (0, 0, 2, 2, 2, \dots)$ and $\frac{1}{4} = Φ (0, 2, 0, 2, \dots)$ , the alternating sequence (Which points of $[0, 1]$ lie in the Cantor set, read off their ternary expansions, with $1 / 4$ worked out).

[L4]

$c (t) = γ (t)$ for $t \in C$ ; $c$ is constant on $[u, v]$ whenever $u < v$ , $u, v \in C$ and $(u, v) \cap C = \emptyset$ (The Cantor function is well defined, satisfies $c (x) \leq c (y)$ whenever $x \leq y$ , is surjective onto $[0, 1]$ , and is constant on every interval removed from the Cantor set, claims 1 and 4).

[L5]

$C \subseteq C_{1} = [0, \frac{1}{3}] \cup [\frac{2}{3}, 1]$ (The Cantor middle-thirds set as the intersection of the sets $C_{n}$ obtained by removing open middle thirds, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

[L6]

Ordered-field arithmetic: $0 < 1$ , so $2 > 0$ , $3 > 0$ , $4 > 0$ and $\frac{1}{3} < \frac{2}{3}$ ; 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

One more digit sequence. $\frac{7}{9} = Φ (2, 0, 2, 2, 2, \dots)$ : by [L2] that value is $2 \cdot 3^{- 1} + 0 + \sum_{k \geq 2} 2 \cdot 3^{- k - 1} = \frac{2}{3} + 3^{- 2} = \frac{2}{3} + \frac{1}{9} = \frac{7}{9}$ .

givenL1L2L6

1.2

The value of $γ$ at the alternating sequence. Let $b$ be the ${0, 1}$ -valued sequence with $b_{k} = 0$ for even $k$ and $b_{k} = 1$ for odd $k$ , and put $B : = \sum_{k \geq 0} b_{k} 2^{- k - 1}$ , which converges by [L2]. Splitting off the first term twice as in [L2] gives $B = 0 \cdot 2^{- 1} + 2^{- 1} B^{'}$ with $B^{'} : = \sum_{k \geq 0} b_{k + 1} 2^{- k - 1}$ , and $B^{'} = 1 \cdot 2^{- 1} + 2^{- 1} B$ , since shifting twice returns $b$ . Hence $B = 2^{- 1} (2^{- 1} + 2^{- 1} B) = 4^{- 1} + 4^{- 1} B$ , so $(1 - 4^{- 1}) B = 4^{- 1}$ and $B = \frac{1}{3}$ by [L6].

givenL2L6

2.1

The five values of $γ$ . By [L1], [L2] and [L3]: $γ (\frac{1}{3}) = \sum_{k \geq 1} 2^{- k - 1} = 2^{- 1}$ ; $γ (\frac{2}{3}) = 1 \cdot 2^{- 1} = 2^{- 1}$ ; $γ (\frac{1}{9}) = \sum_{k \geq 2} 2^{- k - 1} = 2^{- 2} = \frac{1}{4}$ ; $γ (\frac{1}{4}) = B = \frac{1}{3}$ by step 1.2, the halved digits of the alternating ternary sequence being exactly $b$ ; and $γ (\frac{7}{9}) = 2^{- 1} + 0 + \sum_{k \geq 2} 2^{- k - 1} = \frac{1}{2} + \frac{1}{4} = \frac{3}{4}$ by step 1.1.

step 1.1step 1.2L1L2L3L6

3.1

The values of $c$ . All five points lie in $C$ , so $c$ agrees with $γ$ there by [L4]: $c (\frac{1}{9}) = \frac{1}{4}$ , $c (\frac{1}{4}) = \frac{1}{3}$ , $c (\frac{7}{9}) = \frac{3}{4}$ and $c (\frac{1}{3}) = c (\frac{2}{3}) = \frac{1}{2}$ . Moreover $\frac{1}{3} < \frac{2}{3}$ by [L6], both lie in $C$ , and $(\frac{1}{3}, \frac{2}{3}) \cap C = \emptyset$ by [L5]; so [L4] gives that $c$ is constant on $[\frac{1}{3}, \frac{2}{3}]$ , with the value $c (\frac{1}{3}) = \frac{1}{2}$ .

step 2.1L1L3L4L5L6∎

Remarks

The staircase is visible in these five numbers. $c$ rises from $c (\frac{1}{9}) = \frac{1}{4}$ to $c (\frac{1}{3}) = \frac{1}{2}$ across the second stage, stays at $\frac{1}{2}$ across the whole middle third, and reaches $c (\frac{7}{9}) = \frac{3}{4}$ ; between those it is constant on each removed interval (The Cantor function is well defined, satisfies $c (x) \leq c (y)$ whenever $x \leq y$ , is surjective onto $[0, 1]$ , and is constant on every interval removed from the Cantor set). All of the rise happens on $C$ , a set of measure zero (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).
$c (1 / 4) = 1 / 3$ is the value at a non-endpoint. $1 / 4$ lies in $C$ and is the endpoint of no removed interval ( $1 / 4$ lies in the Cantor set and is the endpoint of no removed interval, so the endpoints do not exhaust it), so $c$ is not constant on any neighbourhood of it; the computation of step 1.2 is the only one of the five that cannot be read off a finite digit string.
Halving digits is a bijection, not an approximation. The value $γ (x)$ is the exact sum of the halved-digit series, and the two series are compared term by term, never numerically; this is why $c (\frac{1}{3})$ and $c (\frac{2}{3})$ come out equal, the sequences $(0, 2, 2, 2, \dots)$ and $(2, 0, 0, 0, \dots)$ halving to $(0, 1, 1, 1, \dots)$ and $(1, 0, 0, 0, \dots)$ , both summing to $\frac{1}{2}$ .

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Sources

Cantor function (Wikipedia) (standard reference, not scraped)
Cantor set (Wikipedia) (standard reference, not scraped)

The Cantor function takes the value 1/21/2 on all of [1/3,2/3][1/3,2/3], and its values at 1/91/9, 1/41/4 and 7/97/9