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The Cantor set is homeomorphic to ${0, 1}^{N}$ with the product of discrete topologies, the ternary digits being the coordinates

Example

Let $D : = {0, 1}$ carry the discrete topology (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies) and let

$K : = D^{N} = \prod_{j \in N} D$

carry the product topology (The product set $\prod_{i \in I} X_{i}$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space). Let $C$ be the Cantor set (The Cantor middle-thirds set as the intersection of the sets $C_{n}$ obtained by removing open middle thirds) with the subspace topology inherited from the usual topology of $R$ (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace, The absolute value makes $R$ a metric space: $d (x, y) = ∣ x - y ∣$ is a metric, its open balls are the intervals $(x - r, x + r)$ , and it is unbounded). Define

$Φ : K \to C, Φ (b) : = \sum_{j = 0}^{\infty} 2 b_{j} 3^{- (j + 1)} .$

Then:

$Φ$ is a well defined bijection onto $C$ . Writing $Φ_{0} (a) : = \sum_{k = 0}^{\infty} a_{k} 3^{- k - 1}$ for a sequence $a : N \to {0, 2}$ , one has $Φ (b) = Φ_{0} ((2 b_{k})_{k})$ , and claim 3 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 ${0, 1}^{N}$ says exactly that this assignment is a bijection from ${0, 1}^{N}$ onto $C$ .
Two estimates control $Φ$ completely. For $b, c \in K$ $b, c \in K$ and $n \in N$ $n \in N$ :
- if $b_{j} = c_{j}$ for every $j < n$ , then $∣ Φ (b) - Φ (c) ∣ \leq 3^{- n}$ ;
- if $b_{j} = c_{j}$ for every $j < m$ and $b_{m} \neq c_{m}$ , then $∣ Φ (b) - Φ (c) ∣ \geq 3^{- (m + 1)}$ .
$Φ$ is a homeomorphism (Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological): it is continuous by the first estimate and open onto $C$ by the second, and a continuous open bijection is a homeomorphism (A continuous bijection is a homeomorphism iff it is open iff it is closed, and homeomorphy is an equivalence relation on spaces).

So the Cantor set, a subspace of the line, and the space of all binary sequences, a product of two-point discrete spaces, are the same topological space; the ternary digits of a point of $C$ are its coordinates in the product.

Compactness is not used anywhere below. The usual argument, that a continuous bijection from a compact space to a Hausdorff space is a homeomorphism, is not available at this point in the reading order, and the openness of $Φ$ is proved by hand instead.

Facts & Assumptions

Given: $D = {0, 1}$ discrete, $K = D^{N}$ with the product topology, the Cantor set $C \subseteq R$ with the subspace topology, the map $Φ$ above, and points $b, c \in K$ . For $n \in N$ the cylinder at $b$ of depth $n$ is $Z (b, n) : = {c \in K : c_{j} = b_{j} for every j < n}$ . Powers $3^{- k}$ are integer powers (Integer powers $a^{m}$ ) and $3$ denotes $ι (3)$ (The canonical natural $ι (n) = n \cdot 1_{F}$ of a field).

[A1]

For a sequence $a : N \to {0, 2}$ the series $\sum_{k \geq 0} a_{k} 3^{- k - 1}$ converges, its sum $Φ_{0} (a)$ lies in $[0, 1]$ , the Cantor set is exactly the set of these sums, and $b \mapsto Φ_{0} ((2 b_{k})_{k})$ is a bijection from ${0, 1}^{N}$ onto $C$ (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 ${0, 1}^{N}$ , claims 1, 2 and 3; The Cantor middle-thirds set as the intersection of the sets $C_{n}$ obtained by removing open middle thirds).

[A2]

A basis for the product topology on $K$ is the family of boxes $\prod_{j} U_{j}$ with every $U_{j} \subseteq D$ and $U_{j} = D$ off a list $j_{0}, \dots, j_{n - 1}$ ; every subset of $D$ is open (The product set $\prod_{i \in I} X_{i}$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space, The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies, A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice).

[A3]

$C$ is metrizable, being a subspace of the metrizable space $R$ , and its topology is the metric topology of the restricted metric, whose balls are $B_{C} (x, r) = (x - r, x + r) \cap C$ (Every subspace of a metrizable space is metrizable and every subspace of a first countable space is first countable, the metric case being the subspace metric already identified with the subspace topology, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not, The absolute value makes $R$ a metric space: $d (x, y) = ∣ x - y ∣$ is a metric, its open balls are the intervals $(x - r, x + r)$ , and it is unbounded, Open ball, closed ball and sphere in a metric space, The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, For $n \geq 1$ the product topology on $n$ copies of the usual topology of $R$ is the metric topology of $d_{\infty}$ on $R^{n}$ , and hence also of $d_{1}$ and $d_{2}$ , so $R^{n}$ as a product and $R^{n}$ as a metric space are one space).

[L1]

$\sum_{k = 0}^{\infty} r^{k} = 1 / (1 - r)$ for $∣ r ∣ < 1$ (For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges); a nonnegative series converges iff its partial sums are bounded, and then every partial sum is at most the sum (A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum); series may be shifted to a general starting index (Series, partial sums, convergence and the sum, divergence, and the tail series); sums are additive and homogeneous (Convergent series add and scale termwise).

[L2]

Finite sums are monotone in their terms and satisfy $\sum_{k < n} λ = n λ$ (Laws of finite sums and finite products); weak inequalities pass to limits (Limits preserve non-strict inequalities); $∣ u ∣ \leq v$ is equivalent to $- v \leq u \leq v$ (Basic properties of the absolute value).

[L3]

$3^{m} 3^{n} = 3^{m + n}$ and $3^{- m} = (3^{m})^{- 1}$ (Laws of integer exponents, Integer powers $a^{m}$ ); $3^{k} > 0$ and $k \mapsto 3^{k}$ is nondecreasing for $k \in N$ , so $0 < u \leq v$ gives $3^{- v} \leq 3^{- u}$ (Inverses of positives are positive, and reciprocation reverses order, Laws of integer exponents).

[L4]

If $\sum a_{k}$ converges then $a_{k} \to 0$ (If a series converges then its terms tend to $0$ ); below any positive real lies a positive rational, so convergence tested against rational tolerances gives every real tolerance (The rationals embed densely in the reals).

[L5]

Every nonempty set of naturals has a least element (The well-ordering principle); a listed finite set of reals has a maximum (Every nonempty finite set of reals has a maximum and a minimum).

[L6]

A continuous bijection is a homeomorphism exactly when it is an open map (A continuous bijection is a homeomorphism iff it is open iff it is closed, and homeomorphy is an equivalence relation on spaces, claim 1; Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological).

Verification

technique · direct

1.1

For every $b \in K$ the terms $2 b_{j} 3^{- (j + 1)}$ lie between $0$ and $2 \cdot 3^{- (j + 1)}$ , and $\sum_{j} 2 \cdot 3^{- (j + 1)}$ converges with sum $1$ , since it is $2 \cdot 3^{- 1} \sum_{j} 3^{- j} = (2 / 3) (3 / 2)$ by [L1] and [L3]. So $Φ (b)$ is defined and lies in $[0, 1]$ .

givenL1L2L3

1.2

$3^{- k} \to 0$ , since $\sum_{k} 3^{- k}$ converges by [L1]; hence for every real $ε > 0$ there is $n \in N$ with $3^{- n} < ε$ , a positive rational below $ε$ serving as the tolerance.

L1L4

1.3

Each cylinder $Z (b, n)$ is open in $K$ : it is the box with factor ${b_{j}}$ at each $j < n$ and $D$ elsewhere, and every subset of $D$ is open. Moreover $Z (c, n) = Z (b, n)$ whenever $c \in Z (b, n)$ , the defining condition being agreement of the first $n$ coordinates.

1.4

The cylinders form a basis of $K$ : given a basic box $\prod_{j} U_{j}$ with $U_{j} = D$ off a list $j_{0}, \dots, j_{r - 1}$ and a point $b$ in it, put $n : = 1 + j_{m_{0}}$ where $j_{m_{0}}$ is the largest of the listed indices, available by [L5] for $r \geq 1$ , and $n : = 0$ for $r = 0$ ; then $b \in Z (b, n) \subseteq \prod_{j} U_{j}$ , since every listed index is $< n$ .

A2L5

2.1

$Φ (b) = Φ_{0} ((2 b_{k})_{k})$ for every $b \in K$ , the two series having the same terms $2 b_{k} 3^{- k - 1}$ ; so by [A1] the map $Φ$ is a bijection of $K$ onto $C$ . This is claim 1.

step 1.1A1L1

2.2

For every $n \in N$ : $\sum_{j \geq n} 2 \cdot 3^{- (j + 1)} = 3^{- n}$ , by shifting the index and applying [L1] and [L3] as in step 1.1.

step 1.1L1L3

3.1

Suppose $b_{j} = c_{j}$ for every $j < n$ . For $N \geq n$ the finite sum $\sum_{j < N} 2 (b_{j} - c_{j}) 3^{- (j + 1)}$ has vanishing terms for $j < n$ , and each remaining term lies between $- 2 \cdot 3^{- (j + 1)}$ and $2 \cdot 3^{- (j + 1)}$ , so by [L2] the finite sum lies between $- 3^{- n}$ and $3^{- n}$ , using step 2.2 and [L1]. Letting $N$ grow and applying [L1] and [L2] gives $∣ Φ (b) - Φ (c) ∣ \leq 3^{- n}$ .

step 2.2L1L2

3.2

Suppose $b_{j} = c_{j}$ for every $j < m$ and $b_{m} \neq c_{m}$ ; interchanging $b$ and $c$ if necessary, take $b_{m} = 1$ and $c_{m} = 0$ . For $N > m$ the finite sum $\sum_{j < N} 2 (b_{j} - c_{j}) 3^{- (j + 1)}$ equals $2 \cdot 3^{- (m + 1)}$ plus a term bounded below by $- \sum_{j \geq m + 1} 2 \cdot 3^{- (j + 1)} = - 3^{- (m + 1)}$ , by step 2.2, [L1] and [L2]; so it is at least $3^{- (m + 1)}$ . Letting $N$ grow gives $Φ (b) - Φ (c) \geq 3^{- (m + 1)}$ , hence $∣ Φ (b) - Φ (c) ∣ \geq 3^{- (m + 1)}$ .

step 2.2L1L2

4.1

Steps 3.1 and 3.2 are claim 2.

step 3.1step 3.2

4.2

$Φ$ is continuous: let $V$ be open in $C$ and $b \in Φ^{- 1} [V]$ ; by [A3] there is $ε > 0$ with $B_{C} (Φ (b), ε) \subseteq V$ , by step 1.2 there is $n$ with $3^{- n} < ε$ , and by step 3.1 every $c \in Z (b, n)$ has $∣ Φ (c) - Φ (b) ∣ \leq 3^{- n} < ε$ , so $Z (b, n) \subseteq Φ^{- 1} [V]$ ; and $Z (b, n)$ is open by step 1.3.

step 1.2step 1.3step 3.1A3

4.3

For $b \in K$ and $n \in N$ : $B_{C} (Φ (b), 3^{- n}) \subseteq Φ [Z (b, n)]$ . Indeed such a point is $Φ (c)$ for a unique $c \in K$ by step 2.1; if $c \notin Z (b, n)$ , let $m$ be the least index with $b_{m} \neq c_{m}$ , which exists by [L5] and satisfies $m < n$ , and then step 3.2 gives $∣ Φ (c) - Φ (b) ∣ \geq 3^{- (m + 1)} \geq 3^{- n}$ by [L3], contradicting the choice of $Φ (c)$ .

step 2.1step 3.2L3L5

5.1

$Φ$ is an open map onto $C$ : by step 1.3 the set $Φ [Z (b, n)]$ contains, around each of its points $Φ (c)$ with $c \in Z (b, n)$ , the ball $B_{C} (Φ (c), 3^{- n}) \subseteq Φ [Z (c, n)] = Φ [Z (b, n)]$ by step 4.3; so each $Φ [Z (b, n)]$ is open in $C$ by [A3], and by step 1.4 every open subset of $K$ is a union of cylinders, whose image is the union of their images.

step 1.3step 1.4step 4.3A3

6.1

By step 2.1 the map $Φ$ is a bijection onto $C$ , by step 4.2 it is continuous and by step 5.1 it is open, so it is a homeomorphism by [L6]. This is claim 3, and with steps 2.1 and 4.1 all three claims are proved.

step 2.1step 4.1step 4.2step 5.1L6∎

Remarks

The two estimates say that $Φ$ almost preserves distance. Agreement of the first $n$ coordinates forces the images to be within $3^{- n}$ , and the first disagreement at index $m$ forces them to be at least $3^{- (m + 1)}$ apart. Together they say that the cylinder $Z (b, n)$ and the trace on $C$ of an interval of length about $3^{- n}$ around $Φ (b)$ determine each other, which is exactly what makes $Φ$ a homeomorphism.
Why openness has to be proved and not quoted. For a continuous bijection, openness is equivalent to being a homeomorphism (A continuous bijection is a homeomorphism iff it is open iff it is closed, and homeomorphy is an equivalence relation on spaces) and is not automatic; the standard shortcut uses compactness of the source and the Hausdorff condition on the target, and compactness is later in the reading order. Steps 4.3 and 5.1 replace it with a direct computation.
The coordinates are the digits, and the digits are not the point. A real number in $C$ has exactly one ternary expansion with digits in ${0, 2}$ , which is what makes $Φ$ injective; the ambiguity of ternary expansions in general, such as two expansions of $1 / 3$ , does not arise inside $C$ because the alternative expansion uses the digit $1$ .

Depends on

Used by

Nothing in the library uses this result yet.

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Sources

Cantor set (Wikipedia) (standard reference, not scraped)
Cantor space (Wikipedia) (standard reference, not scraped)
Product topology (Wikipedia) (standard reference, not scraped)

The Cantor set is homeomorphic to {0,1}N{0,1}N with the product of discrete topologies, the ternary digits being the coordinates