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Which points of $[0,1]$ lie in the Cantor set, read off their ternary expansions, with $1/4$ worked out

Example

By 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 $\{0,1\}^{\mathbb{N}}$ a real $x$ lies in the Cantor set $C$ exactly when

$x \;=\; \Phi(a) \;=\; \sum_{k=0}^{\infty} a_k 3^{-k-1} \qquad \text{for some sequence } a \text{ with every } a_k \in \{0,2\},$

and that sequence is then unique. The membership test is therefore: $x$ has a ternary expansion using only the digits $0$ and $2$ . Six points are worked out here.

$x$	digit sequence $\Phi^{-1}(x)$
$0$	$0,0,0,0,\dots$
$1$	$2,2,2,2,\dots$
$1/3$	$0,2,2,2,\dots$
$2/3$	$2,0,0,0,\dots$
$1/9$	$0,0,2,2,2,\dots$
$1/4$	$0,2,0,2,0,2,\dots$

The last line is the interesting one: the digits of $1/4$ alternate for ever, so $1/4$ lies in $C$ without being an endpoint of any interval removed in the construction ( $1/4$ lies in the Cantor set and is the endpoint of no removed interval, so the endpoints do not exhaust it).

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 $\{0,1\}^{\mathbb{N}}$ . Write $\sigma a$ for the shifted sequence $(\sigma a)_k = a_{k+1}$ .

[L1]

$\Phi$ is a bijection from $D$ onto $C$ with $\Phi(a) = \sum_{k \ge 0} a_k 3^{-k-1}$ , and the series converges for every $a \in D$ (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 $\{0,1\}^{\mathbb{N}}$ , Series, partial sums, convergence and the sum, divergence, and the tail series, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L2]

$\sum_{k=0}^{\infty} r^{k} = 1/(1-r)$ for $|r| < 1$ ; in particular $\sum_{k \ge 0} 3^{-k} = 3 \cdot 2^{-1}$ and hence $\sum_{k \ge m} 2 \cdot 3^{-k-1} = 3^{-m}$ ; convergent series add and scale termwise, and the tail of a convergent series is again convergent with $\sum_{k \ge 0} t_k = t_0 + \sum_{k \ge 1} t_k$ (For $|r| < 1$ , $\sum_{k \ge 0} r^k = 1/(1-r)$ , and for $|r| \ge 1$ 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).

[L4]

Ordered-field arithmetic: $0 < 1$ , so $3 > 0$ , $3^{-1} > 0$ , $8 > 0$ ; adding a constant and multiplying by a positive preserve an inequality; the order is total and transitive (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

The shift identity. For $a \in D$ , $\Phi(a) = a_0 \cdot 3^{-1} + 3^{-1}\Phi(\sigma a)$ : by [L2] the series splits as $\Phi(a) = a_0 3^{-1} + \sum_{k \ge 1} a_k 3^{-k-1}$ , and $\sum_{k \ge 1}a_k 3^{-k-1} = \sum_{j \ge 0} a_{j+1} 3^{-j-2} = 3^{-1}\sum_{j \ge 0}a_{j+1}3^{-j-1} = 3^{-1}\Phi(\sigma a)$ by [L2] and [L4].

givenL1L2L4

1.2

The constant and eventually constant sequences. By [L2], $\Phi(0,0,0,\dots) = 0$ and $\Phi(2,2,2,\dots) = \sum_{k \ge 0}2 \cdot 3^{-k-1} = 3^{0} = 1$ . Likewise $\Phi(0,2,2,2,\dots) = 0 + \sum_{k \ge 1}2\cdot3^{-k-1} = 3^{-1} = 1/3$ , $\Phi(2,0,0,\dots) = 2 \cdot 3^{-1} = 2/3$ , and $\Phi(0,0,2,2,2,\dots) = \sum_{k \ge 2}2 \cdot 3^{-k-1} = 3^{-2} = 1/9$ .

L1L2

2.1

The alternating sequence gives $1/4$ . Let $a$ be the sequence with $a_k = 0$ for even $k$ and $a_k = 2$ for odd $k$ , so $a \in D$ and $\sigma(\sigma a) = a$ . Applying step 1.1 twice, $\Phi(a) = 0 \cdot 3^{-1} + 3^{-1}\Phi(\sigma a)$ and $\Phi(\sigma a) = 2 \cdot 3^{-1} + 3^{-1}\Phi(a)$ , so $\Phi(a) = 3^{-1}\big(2 \cdot 3^{-1} + 3^{-1}\Phi(a)\big) = 2 \cdot 9^{-1} + 9^{-1}\Phi(a)$ ; hence $(1 - 9^{-1})\Phi(a) = 2 \cdot 9^{-1}$ , that is $8 \cdot 9^{-1}\Phi(a) = 2 \cdot 9^{-1}$ and $\Phi(a) = 2 \cdot 8^{-1} = 1/4$ , by [L4].

step 1.1L4

3.1

So all six points of the table lie in $C$ by [L1], with the digit sequences shown, and the sequences are the only ones representing them because $\Phi$ is injective by [L1]. The point $1/4$ has a digit sequence that is not eventually constant, since it takes both values $0$ and $2$ at arbitrarily large indices.

step 1.2step 2.1L1∎

Remarks

The digit $1$ is what the test forbids. $1/3$ has the ternary expansion $0.1000\dots$ as well as $0.0222\dots$ , and it is the second that witnesses $1/3 \in C$ ; the test asks for the existence of an expansion with digits in $\{0,2\}$ , not for every expansion to have that form. By contrast $1/2$ is not in $C$ at all, since $C \subseteq C_1 = [0,\tfrac13] \cup [\tfrac23,1]$ (The Cantor middle-thirds set as the intersection of the sets $C_n$ obtained by removing open middle thirds) while $\tfrac13 < \tfrac12 < \tfrac23$ .
The construction and the digits match stage by stage. $C_1$ keeps the points whose first digit can be taken $0$ or $2$ , $C_2$ those whose first two digits can be, and so on; that correspondence is the content 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 $\{0,1\}^{\mathbb{N}}$ and is what makes the table computable without ever drawing the intervals.
$1/4$ is not special. Every point of $C$ whose digit sequence is not eventually constant fails to be an endpoint, and those points are the vast majority: the eventually constant sequences are at most countable while $C$ is not (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).

Depends on

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Sources

Cantor set (Wikipedia) (standard reference, not scraped)
Ternary numeral system (Wikipedia) (standard reference, not scraped)
Stanford Math 205A, Homework 1 (standard reference, not scraped)

Which points of [0,1][0,1][0,1] lie in the Cantor set, read off their ternary expansions, with 1/41/41/4 worked out