Editing Cardinality of the continuum (section)

===Cardinal equalities===
A variation of Cantor's diagonal argument can be used to prove [[Cantor's theorem]], which states that the cardinality of any set is strictly less than that of its [[power set]]. That is, <math>|A| < 2^{|A|}</math> (and so that the power set <math>\wp(\mathbb N)</math> of the [[natural number]]s <math>\mathbb N</math> is uncountable).<ref>{{SpringerEOM|title=Cantor theorem|oldid=51494|mode=cs1}}</ref> In fact, the cardinality of <math>\wp(\mathbb N)</math>, by definition <math>2^{\aleph_0}</math>, is equal to <math>{\mathfrak c}</math>. This can be shown by providing one-to-one mappings in both directions between subsets of a countably infinite set and real numbers, and applying the [[Cantor–Bernstein–Schroeder theorem]] according to which two sets with one-to-one mappings in both directions have the same cardinality.<ref name=stillwell>{{cite journal
 | last = Stillwell | first = John
 | doi = 10.1080/00029890.2002.11919865
 | issue = 3
 | journal = American Mathematical Monthly
 | jstor = 2695360
 | mr = 1903582
 | pages = 286–297
 | title = The continuum problem
 | volume = 109
 | year = 2002}}</ref><ref name=johnson>{{cite book
 | last = Johnson | first = D. L.
 | department = Elements of Logic via Numbers and Sets
 | doi = 10.1007/978-1-4471-0603-6_6
 | isbn = 9781447106036
 | series = Springer Undergraduate Mathematics Series
 | pages = 113–130
 | publisher = Springer London
 | title = Chapter 6: Cardinal numbers
 | chapter = Cardinal Numbers
 | year = 1998}}</ref> In one direction, reals can be equated with [[Dedekind cut]]s, sets of rational numbers,<ref name=stillwell/> or with their [[binary expansion]]s.<ref name=johnson/> In the other direction, the binary expansions of numbers in the half-open interval <math>[0,1)</math>, viewed as sets of positions where the expansion is one, almost give a one-to-one mapping from subsets of a countable set (the set of positions in the expansions) to real numbers, but it fails to be one-to-one for numbers with terminating binary expansions, which can also be represented by a non-terminating expansion that ends in a repeating sequence of 1s. This can be made into a one-to-one mapping by that adds one to the non-terminating repeating-1 expansions, mapping them into <math>[1,2)</math>.<ref name=johnson/> Thus, we conclude that<ref name=stillwell/><ref name=johnson/>
{{block indent|<math>\mathfrak c = |\wp(\mathbb{N})| = 2^{\aleph_0}.</math>}}

The cardinal equality <math>\mathfrak{c}^2 = \mathfrak{c}</math> can be demonstrated using [[cardinal arithmetic]]:
{{block indent|<math>\mathfrak{c}^2 = (2^{\aleph_0})^2 = 2^{2\times{\aleph_0}} = 2^{\aleph_0} = \mathfrak{c}.</math>}}

By using the rules of cardinal arithmetic, one can also show that

{{block indent|<math>\mathfrak c^{\aleph_0} = {\aleph_0}^{\aleph_0} = n^{\aleph_0} = \mathfrak c^n = \aleph_0 \mathfrak c = n \mathfrak c = \mathfrak c</math>}}

where ''n'' is any finite cardinal ≥ 2 and

{{block indent|<math> \mathfrak c ^{\mathfrak c}  =  (2^{\aleph_0})^{\mathfrak c}  = 2^{\mathfrak c\times\aleph_0} = 2^{\mathfrak c}</math>}}

where <math>2 ^{\mathfrak c}</math> is the cardinality of the power set of '''R''', and <math>2 ^{\mathfrak c} > \mathfrak c </math>.