CounterexampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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$1/4$ lies in the Cantor set and is the endpoint of no removed interval, so the endpoints do not exhaust it

Statement refuted

Refuted claim: the Cantor set $C$ consists of the endpoints of the removed intervals and is therefore at most countable (FALSE: the Cantor set is countable because only countably many intervals were removed).

The witness is $x_0 := 1/4$ . It lies in $C$ , its ternary digit sequence being the alternating sequence $(0,2,0,2,\dots)$ (Which points of $[0,1]$ lie in the Cantor set, read off their ternary expansions, with $1/4$ worked out), and it is the endpoint of no interval removed in the construction. Here, as everywhere on this page, $u$ is an endpoint of a removed interval means that $u \in C$ and there is $v \ne u$ with the open interval between $u$ and $v$ disjoint from $C$ ; that is exactly what "the interval between them was removed" says in the vocabulary available. What is shown below is that $C$ meets every interval $(x_0, x_0 + \delta)$ and every interval $(x_0 - \delta, x_0)$ , for every real $\delta > 0$ , so no such $v$ exists on either side.

Facts & Assumptions

Given: The Cantor set $C$ , the set $D$ of $\{0,2\}$ -valued sequences and the bijection $\Phi : D \to C$ of The Cantor set is exactly the set of $\sum_{k \ge 1} a_k 3^{-k}$ with every $a_k \in \{0,2\}$ , and this gives a bijection with $\{0,1\}^{\mathbb{N}}$ ; the alternating sequence $a \in D$ , with $a_j = 0$ for even $j$ and $a_j = 2$ for odd $j$ ; and $x_0 := \Phi(a)$ .

[A1]

The refuted claim: every point of $C$ is an endpoint of a removed interval, so $C$ is at most countable.

[L1]

$\Phi$ is a bijection from $D$ onto $C$ with $\Phi(b) = \sum_{j \ge 0}b_j 3^{-j-1}$ , and $x_0 = \Phi(a) = 1/4$ (The Cantor set is exactly the set of $\sum_{k \ge 1} a_k 3^{-k}$ with every $a_k \in \{0,2\}$ , and this gives a bijection with $\{0,1\}^{\mathbb{N}}$ , Which points of $[0,1]$ lie in the Cantor set, read off their ternary expansions, with $1/4$ worked out, The Cantor middle-thirds set as the intersection of the sets $C_n$ obtained by removing open middle thirds).

[L2]

Geometric tails: $\sum_{j \ge m}2 \cdot 3^{-j-1} = 3^{-m}$ ; a series of nonnegative terms has nonnegative sum and all partial sums at most the sum; convergent series add and scale termwise; and a series splits as $\sum_{j \ge 0} t_j = \sum_{j<m} t_j + \sum_{j \ge m} t_j$ (For $|r| < 1$ , $\sum_{k \ge 0} r^k = 1/(1-r)$ , and for $|r| \ge 1$ the series diverges, A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum, Convergent series add and scale termwise, Series, partial sums, convergence and the sum, divergence, and the tail series, Integer powers $a^m$ , Laws of integer exponents).

[L3]

$3^{-n} \to 0$ , and convergence to $0$ is tested against rational $\varepsilon > 0$ (For $|r| < 1$ the sequence $r^k$ is null, and for $|r| > 1$ the sequence $|r|^k$ diverges to $+\infty$ , Limits and Cauchy sequences of reals).

[L4]

Ordered-field arithmetic: $0 < 1$ , so $3 > 0$ and $3^{-1} > 0$ and $2 \cdot 3^{-1} < 1$ ; $3^{-p} \le 3^{-q}$ whenever $q \le p$ , by induction from $0 < 3^{-1} < 1$ ; 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), Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length). 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.

Counterexample

technique · direct

1.1

By [L1] the point $x_0 = 1/4$ lies in $C$ and has digit sequence $a$ , with $a_{2m} = 0$ and $a_{2m+1} = 2$ for every $m \in \mathbb{N}$ . By [L2], for every $m$ one may split $\Phi(a) = \sum_{j<2m}a_j3^{-j-1} + \sum_{j \ge 2m}a_j3^{-j-1}$ , and every tail $\sum_{j \ge p}a_j3^{-j-1}$ lies between $0$ and $3^{-p}$ .

givenL1L2

2.1

Points of $C$ immediately above $x_0$ . For $m \in \mathbb{N}$ let $b \in D$ agree with $a$ at every index $< 2m$ , have $b_{2m} = 2$ , and have $b_j = 0$ for $j > 2m$ . Writing $P := \sum_{j<2m}a_j3^{-j-1}$ , step 1.1 and [L2] give $\Phi(b) = P + 2 \cdot 3^{-2m-1}$ and $\Phi(a) = P + 0 + T$ with $0 \le T \le 3^{-2m-1}$ , so $\Phi(b) - x_0 = 2 \cdot 3^{-2m-1} - T$ lies between $3^{-2m-1}$ and $2 \cdot 3^{-2m-1}$ ; in particular $\Phi(b) > x_0$ and $\Phi(b) - x_0 \le 2 \cdot 3^{-2m-1} < 3^{-2m} \le 3^{-m}$ by [L4]. And $\Phi(b) \in C$ by [L1].

step 1.1L1L2L4

2.2

Points of $C$ immediately below $x_0$ . For $m \in \mathbb{N}$ let $d \in D$ agree with $a$ at every index $< 2m+1$ , have $d_{2m+1} = 0$ , and have $d_j = 2$ for $j > 2m+1$ . Writing $Q := \sum_{j<2m+1}a_j3^{-j-1}$ , step 1.1 and [L2] give $\Phi(d) = Q + 0 + 3^{-2m-2}$ and $\Phi(a) = Q + 2 \cdot 3^{-2m-2} + T'$ with $0 \le T' \le 3^{-2m-2}$ , so $x_0 - \Phi(d) = 3^{-2m-2} + T'$ lies between $3^{-2m-2}$ and $2 \cdot 3^{-2m-2}$ ; in particular $\Phi(d) < x_0$ and $x_0 - \Phi(d) < 3^{-2m-1} \le 3^{-m}$ by [L4]. And $\Phi(d) \in C$ by [L1].

step 1.1L1L2L4

3.1

Let the real $\delta > 0$ be given; by [L3] fix $m$ with $3^{-m} < \delta$ . Steps 2.1 and 2.2 then produce points of $C$ in $(x_0,\ x_0 + \delta)$ and in $(x_0 - \delta,\ x_0)$ . Consequently, for every $v > x_0$ the interval $(x_0, v)$ meets $C$ , and for every $u < x_0$ the interval $(u, x_0)$ meets $C$ ; so there is no $v \ne x_0$ with the open interval between $x_0$ and $v$ disjoint from $C$ , and $x_0$ is the endpoint of no removed interval. Since $x_0 \in C$ , the claim [A1] fails at $x_0$ .

step 2.1step 2.2A1L1L3L4∎

Remarks

The digits diagnose it. By the argument of The Cantor function is well defined, satisfies $c(x) \le c(y)$ whenever $x \le y$ , is surjective onto $[0,1]$ , and is constant on every interval removed from the Cantor set the two endpoints of a gap have digit sequences that are eventually $2$ and eventually $0$ respectively; the digits of $1/4$ alternate for ever, so it can be neither. The proof above avoids that route and exhibits the approximating points directly, which is what makes it self-contained.
How many such points there are. The eventually constant sequences form an at most countable set, while $C$ is uncountable (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 the endpoints are a vanishing part of $C$ and the refuted claim fails not marginally but completely.
$1/4$ is also where the Cantor function takes the value $1/3$ (The Cantor function takes the value $1/2$ on all of $[1/3, 2/3]$ , and its values at $1/9$ , $1/4$ and $7/9$ ), and it is the one value in that example whose computation needs the whole infinite digit sequence rather than a finite initial segment.

Depends on

Used by

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Sources

Cantor set (Wikipedia) (standard reference, not scraped)
Ternary numeral system (Wikipedia) (standard reference, not scraped)

1/41/41/4 lies in the Cantor set and is the endpoint of no removed interval, so the endpoints do not exhaust it