Editing Cantor set (section)

=== 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 set|uncountable]]. To see this, we show that there is a [[function (mathematics)|function]] ''f'' from the Cantor set <math>\mathcal{C}</math> to the closed interval <math>[0, 1]</math> that is [[Surjective function|surjective]] (i.e. ''f'' maps from <math>\mathcal{C}</math> onto <math>[0, 1]</math>) so that the cardinality of <math>\mathcal{C}</math> is no less than that of <math>[0, 1]</math>.  Since <math>\mathcal{C}</math> is a [[subset]] of <math>[0, 1]</math>, 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 <math>[0, 1]</math> interval in terms of base 3 (or [[ternary numeral system|ternary]]) notation. Recall that the proper ternary fractions, more precisely: the elements of <math>\bigl(\Z \setminus \{0\}\bigr) \cdot 3^{-\N_0}</math>, admit more than one representation in this notation, as for example {{sfrac|1|3}}, that can be written as 0.1<sub>3</sub> = {{overline|0.1|0}}<sub>3</sub>, but also as 0.0222...<sub>3</sub> = {{overline|0.0|2}}<sub>3</sub>, and {{sfrac|2|3}}, that can be written as 0.2<sub>3</sub> = {{overline|0.2|0}}<sub>3</sub> but also as 0.1222...<sub>3</sub> = {{overline|0.1|2}}<sub>3</sub>.<ref>This alternative recurring representation of a number with a terminating numeral occurs in any [[Numeral system#Positional systems in detail|positional system]] with [[Absolute value (algebra)#Types of absolute value|Archimedean absolute value]].</ref>
When we remove the middle third, this contains the numbers with ternary numerals of the form 0.1xxxxx...<sub>3</sub> where xxxxx...<sub>3</sub> is strictly between 00000...<sub>3</sub> and 22222...<sub>3</sub>. So the numbers remaining after the first step consist of
* Numbers of the form 0.0xxxxx...<sub>3</sub> (including 0.022222...<sub>3</sub> = 1/3)
* Numbers of the form 0.2xxxxx...<sub>3</sub> (including 0.222222...<sub>3</sub> = 1)

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...<sub>3</sub> and 0.21xxxx...<sub>3</sub>, 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, {{sfrac|1|3}} = 0.1<sub>3</sub> and {{sfrac|7|9}} = 0.21<sub>3</sub> are in the Cantor set, as they have ternary numerals consisting entirely of 0s and 2s: 1 = 0.222...<sub>3</sub> = {{overline|0.|2}}<sub>3</sub>, {{sfrac|1|3}} = 0.0222...<sub>3</sub> = {{overline|0.0|2}}<sub>3</sub> and {{sfrac|7|9}} = 0.20222...<sub>3</sub> = {{overline|0.20|2}}<sub>3</sub>.
All the latter numbers are "endpoints", and these examples are right [[limit point]]s of <math>\mathcal{C}</math>. The same is true for the left limit points of <math>\mathcal{C}</math>, e.g. {{sfrac|2|3}} = 0.1222...<sub>3</sub> = {{overline|0.1|2}}<sub>3</sub> = {{overline|0.2|0}}<sub>3</sub> and {{sfrac|8|9}} = 0.21222...<sub>3</sub> = {{overline|0.21|2}}<sub>3</sub> = {{overline|0.22|0}}<sub>3</sub>. All these endpoints are ''proper ternary'' fractions (elements of <math>\Z \cdot 3^{-\N_0}</math>) of the form {{sfrac|''p''|''q''}}, where denominator ''q'' is a [[power of 3]] when the fraction is in its [[Irreducible fraction|irreducible]] form.<ref name="College"/> 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 <math>\mathcal{C}</math> 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 <math>\mathcal{C}</math> if it again its ternary expansion contains no 1's and "ends" in infinitely many recurring 2s.

This set of endpoints is [[dense set|dense]] in <math>\mathcal{C}</math> (but not dense in <math>[0, 1]</math>) and makes up a [[countably infinite]] set. The numbers in <math>\mathcal{C}</math> 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 <math>\mathcal{C}</math> to <math>[0, 1]</math> 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 numeral system#Representing real numbers|binary]] representation of a real number.  In a formula,

:<math>f \bigg( \sum_{k\in \N} a_k 3^{-k} \bigg) = \sum_{k\in \N} \frac{a_k}{2} 2^{-k}</math> &nbsp; where &nbsp; <math>\forall k\in \N : a_k \in \{0,2\} .</math>

For any number ''y'' in <math>[0, 1]</math>, its binary representation can be translated into a ternary representation of a number ''x'' in <math>\mathcal{C}</math> by replacing all the 1s by 2s.  With this, ''f''(''x'') = ''y'' so that ''y'' is in the [[Range of a function|range]] of ''f''.  For instance if ''y'' = {{sfrac|3|5}} = 0.100110011001...<sub>2</sub> = {{overline|0.|1001}}, we write ''x'' = {{overline|0.|2002}} = 0.200220022002...<sub>3</sub> = {{sfrac|7|10}}.  Consequently, ''f'' is surjective. However, ''f'' is ''not'' [[injective function|injective]] — the values for which ''f''(''x'') coincides are those at opposing ends of one of the ''middle thirds'' removed.  For instance, take
:{{sfrac|1|3}} = {{overline|0.0|2}}<sub>3</sub> (which is a right limit point of <math>\mathcal{C}</math> and a left limit point of the middle third [{{sfrac|1|3}}, {{sfrac|2|3}}])&nbsp; and
:{{sfrac|2|3}} = {{overline|0.2|0}}<sub>3</sub> (which is a left limit point of <math>\mathcal{C}</math> and a right limit point of the middle third [{{sfrac|1|3}}, {{sfrac|2|3}}])
so
:<math>\begin{array}{lcl}
f\bigl({}^1\!\!/\!_3 \bigr) = f(0.0\overline{2}_3) = 0.0\overline{1}_2 = \!\! & \!\! 0.1_2 \!\! & \!\! = 0.1\overline{0}_2 = f(0.2\overline{0}_3) = f\bigl({}^2\!\!/\!_3 \bigr) . \\
& \parallel \\
& {}^1\!\!/\!_2
\end{array}</math>
Thus there are as many points in the Cantor set as there are in the interval <math>[0, 1]</math> (which has the [[Uncountable set|uncountable]] cardinality {{nowrap|<math>\mathfrak{c} = 2^{\aleph_0}</math>).}} 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 {{sfrac|1|4}}, which can be written as 0.020202...<sub>3</sub> = {{overline|0.|02}} in ternary notation.  In fact, given any <math>a\in[-1,1]</math>, there exist <math>x,y\in\mathcal{C}</math> such that <math>a = y-x</math>.  This was first demonstrated by [[Hugo Steinhaus|Steinhaus]] in 1917, who [[mathematical proof|proved]], via a geometric argument, the equivalent assertion that <math>\{(x,y)\in\mathbb{R}^2 \mid y=x+a\} \; \cap \; (\mathcal{C}\times\mathcal{C}) \neq\emptyset</math> for every <math>a\in[-1,1]</math>.<ref>{{Cite book|title=Real Analysis|url=https://archive.org/details/realanalysis00caro_315|url-access=limited|last=Carothers|first=N. L.|publisher=Cambridge University Press|year=2000|isbn=978-0-521-69624-1|location=Cambridge|pages=[https://archive.org/details/realanalysis00caro_315/page/n41 31]–32}}</ref> Since this construction provides an injection from <math>[-1,1]</math> to <math>\mathcal{C}\times\mathcal{C}</math>, we have <math>|\mathcal{C}\times\mathcal{C}|\geq|[-1,1]|=\mathfrak{c}</math> as an immediate [[corollary]].  Assuming that <math>|A\times A|=|A|</math> for any infinite set <math>A</math> (a statement shown to be equivalent to the [[axiom of choice]] [[Tarski's theorem about choice|by Tarski]]), this provides another demonstration that <math>|\mathcal{C}|=\mathfrak{c}</math>.

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 number]]s have the same property, but the Cantor set has the additional property of being [[Closed set|closed]], so it is not even [[Dense set|dense]] in any interval, unlike the irrational numbers which are dense in every interval.

It has been [[conjecture]]d that all [[algebraic number|algebraic]] irrational numbers are [[normal number|normal]].  Since members of the Cantor set are not normal in base 3, this would imply that all members of the Cantor set are either rational or [[transcendental number|transcendental]].