quality:correction, stylistic:tone
Cardinality

It can be shown that there are as many points left behind in this process as there were to begin with, and that therefore, the Cantor set is uncountable. To see this, we show that there is a function f from the Cantor set C {\displaystyle {\mathcal {C}}} to the closed interval that is surjective (i.e. f maps from C {\displaystyle {\mathcal {C}}} onto) so that the cardinality of C {\displaystyle {\mathcal {C}}} is no less than that of. Since C {\displaystyle {\mathcal {C}}} is a subset of, its cardinality is also no greater, so the two cardinalities must in fact be equal, by the Cantor–Bernstein–Schröder theorem.

To construct this function, consider the points in the interval in terms of base 3 (or ternary) notation. Recall that the proper ternary fractions, more precisely: the elements of (Z ∖ { 0 }) ⋅ 3 − N 0 {\displaystyle {\bigl (}\mathbb {Z} \setminus \{0\}{\bigr)}\cdot 3^{-\mathbb {N} _{0}}}, admit more than one representation in this notation, as for example 1 / 3, that can be written as 0.1 = 0.1 0, but also as 0.0222... = 0.0 2, and 2 / 3, that can be written as 0.2 = 0.2 0 but also as 0.1222... = 0.1 2. When we remove the middle third, this contains the numbers with ternary numerals of the form 0.1xxxxx... where xxxxx... is strictly between 00000... and 22222.... So the numbers remaining after the first step consist of

This can be summarized by saying that those numbers with a ternary representation such that the first digit after the radix point is not 1 are the ones remaining after the first step.

The second step removes numbers of the form 0.01xxxx... and 0.21xxxx..., and (with appropriate care for the endpoints) it can be concluded that the remaining numbers are those with a ternary numeral where neither of the first two digits is 1.

Continuing in this way, for a number not to be excluded at step n, it must have a ternary representation whose n th digit is not 1. For a number to be in the Cantor set, it must not be excluded at any step, it must admit a numeral representation consisting entirely of 0s and 2s.

It is worth emphasizing that numbers like 1, 1 / 3 = 0.1 and 7 / 9 = 0.21 are in the Cantor set, as they have ternary numerals consisting entirely of 0s and 2s: 1 = 0.222... = 0. 2, 1 / 3 = 0.0222... = 0.0 2 and 7 / 9 = 0.20222... = 0.20 2.All the latter numbers are "endpoints", and these examples are right limit points of C {\displaystyle {\mathcal {C}}}. The same is true for the left limit points of C {\displaystyle {\mathcal {C}}}, e.g. 2 / 3 = 0.1222... = 0.1 2 = 0.2 0 and 8 / 9 = 0.21222... = 0.21 2 = 0.22 0. All these endpoints are proper ternary fractions (elements of Z ⋅ 3 − N 0 {\displaystyle \mathbb {Z} \cdot 3^{-\mathbb {N} _{0}}}) of the form p / q, where denominator q is a power of 3 when the fraction is in its irreducible form. The ternary representation of these fractions terminates (i.e., is finite) or — recall from above that proper ternary fractions each have 2 representations — is infinite and "ends" in either infinitely many recurring 0s or infinitely many recurring 2s. Such a fraction is a left limit point of C {\displaystyle {\mathcal {C}}} if its ternary representation contains no 1's and "ends" in infinitely many recurring 0s. Similarly, a proper ternary fraction is a right limit point of C {\displaystyle {\mathcal {C}}} if it again its ternary expansion contains no 1's and "ends" in infinitely many recurring 2s.

This set of endpoints is dense in C {\displaystyle {\mathcal {C}}} (but not dense in) and makes up a countably infinite set. The numbers in C {\displaystyle {\mathcal {C}}} which are not endpoints also have only 0s and 2s in their ternary representation, but they cannot end in an infinite repetition of the digit 0, nor of the digit 2, because then it would be an endpoint.

The function from C {\displaystyle {\mathcal {C}}} to is defined by taking the ternary numerals that do consist entirely of 0s and 2s, replacing all the 2s by 1s, and interpreting the sequence as a binary representation of a real number. In a formula,

For any number y in, its binary representation can be translated into a ternary representation of a number x in C {\displaystyle {\mathcal {C}}} by replacing all the 1s by 2s. With this, f (x) = y so that y is in the range of f. For instance if y = 3 / 5 = 0.100110011001... = 0. 1001, we write x = 0. 2002 = 0.200220022002... = 7 / 10. Consequently, f is surjective. However, f is not injective — the values for which f (x) coincides are those at opposing ends of one of the middle thirds removed. For instance, take

so

Thus there are as many points in the Cantor set as there are in the interval (which has the uncountable cardinality c = 2 ℵ 0 {\displaystyle {\mathfrak {c}}=2^{\aleph _{0}}}). However, the set of endpoints of the removed intervals is countable, so there must be uncountably many numbers in the Cantor set which are not interval endpoints. As noted above, one example of such a number is 1 / 4, which can be written as 0.020202... = 0. 02 in ternary notation. In fact, given any a ∈ {\displaystyle a\in }, there exist x, y ∈ C {\displaystyle x,y\in {\mathcal {C}}} such that a = y − x {\displaystyle a=y-x}. This was first demonstrated by Steinhaus in 1917, who proved, via a geometric argument, the equivalent assertion that { (x, y) ∈ R 2 ∣ y = x + a } ∩ (C × C) ≠ ∅ {\displaystyle \{(x,y)\in \mathbb {R} ^{2}\mid y=x+a\}\;\cap \;({\mathcal {C}}\times {\mathcal {C}})\neq \emptyset } for every a ∈ {\displaystyle a\in }. Since this construction provides an injection from {\displaystyle } to C × C {\displaystyle {\mathcal {C}}\times {\mathcal {C}}}, we have | C × C | ≥ | | = c {\displaystyle |{\mathcal {C}}\times {\mathcal {C}}|\geq ||={\mathfrak {c}}} as an immediate corollary. Assuming that | A × A | = | A | {\displaystyle |A\times A|=|A|} for any infinite set A {\displaystyle A} (a statement shown to be equivalent to the axiom of choice by Tarski), this provides another demonstration that | C | = c {\displaystyle |{\mathcal {C}}|={\mathfrak {c}}}.

The Cantor set contains as many points as the interval from which it is taken, yet itself contains no interval of nonzero length. The irrational numbers have the same property, but the Cantor set has the additional property of being closed, so it is not even dense in any interval, unlike the irrational numbers which are dense in every interval.

It has been conjectured that all algebraic irrational numbers are normal. Since members of the Cantor set are not normal, this would imply that all members of the Cantor set are either rational or transcendental.