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{|

s1 >		0, >						| ...)

s2 >			1, >					| ...)

s3 >				0, >				| ...)

s4 >					0, >			| ...)

s5 >						0, >		| ...)

s6 >							1, >	| ...)

s7 >								0, >| ...)

| ...|

s >		1, >	0, >	1, >	1, >	1, >	0, >	1, >| ...)

By construction, s is a member of T that differs from each sn, since their nth digits differ (highlighted in the example).Hence, s cannot occur in the enumeration.Based on this lemma, Cantor then uses a proof by contradiction to show that: The proof starts by assuming that T is countable.Then all its elements can be written in an enumeration s1, s2, ... , sn, ... .Applying the previous lemma to this enumeration produces a sequence s that is a member of T, but is not in the enumeration. However, if T is enumerated, then every member of T, including this s, is in the enumeration. This contradiction implies that the original assumption is false. Therefore, T is uncountable.

Real numbers

The uncountability of the real numbers was already established by Cantor's first uncountability proof, but it also follows from the above result. To prove this, an injection will be constructed from the set T of infinite binary strings to the set R of real numbers. Since T is uncountable, the image of this function, which is a subset of R, is uncountable. Therefore, R is uncountable. Also, by using a method of construction devised by Cantor, a bijection will be constructed between T and R. Therefore, T and R have the same cardinality, which is called the "cardinality of the continuum" and is usually denoted by mathfrak{c} or 2^{aleph_0}.An injection from T to R is given by mapping binary strings in T to decimal fractions, such as mapping t = 0111... to the decimal 0.0111.... This function, defined by {{nowrap|f{{space|hair}}(t) {{=}} 0.t}}, is an injection because it maps different strings to different numbers.While 0.0111... and 0.1000... would be equal if interpreted as binary fractions (destroying injectivity), they are different when interpreted as decimal fractions, as is done by f. On the other hand, since t is a binary string, the equality 0.0999... = 0.1000... of decimal fractions is not relevant here.Constructing a bijection between T and R is slightly more complicated.Instead of mapping 0111... to the decimal 0.0111..., it can be mapped to the base b number: 0.0111...b. This leads to the family of functions: {{nowrap|f'b{{space|hair}}(t) {{=}} 0.t'b}}. The functions {{nowrap|f{{space|hair}}b(t)}} are injections, except for {{nowrap|f{{space|hair}}2(t)}}. This function will be modified to produce a bijection between T and R.{| class="wikitable collapsible collapsed"! Construction of a bijection between T and R style="text-align: left; vertical-align: top"

total_width=200	width1=106	caption1=The function h: (0,1) → (−π/2,Ï€/2)	width2=338	caption2=The function tan: (−π/2,Ï€/2) → R}}This construction uses a method devised by Cantor that was published in 1878. He used it to construct a bijection between the closed interval [0, 1] and the irrationals in the open interval (0, 1). He first removed a countably infinite subset from each of these sets so that there is a bijection between the remaining uncountable sets. Since there is a bijection between the countably infinite subsets that have been removed, combining the two bijections produces a bijection between the original sets.See page 254 of {{Citation|author=Georg Cantor|title=Ein Beitrag zur Mannigfaltigkeitslehre|url=http://www.digizeitschriften.de/dms/img/?PID=GDZPPN002156806|volume=84|pages=242–258|journal=Journal für die Reine und Angewandte Mathematik|year=1878}}. This proof is discussed in {{Citation|author=Joseph Dauben|title=Georg Cantor: His Mathematics and Philosophy of the Infinite|publisher=Harvard University Press|year=1979|isbn=0-674-34871-0}}, pp. 61–62, 65. On page 65, Dauben proves a result that is stronger than Cantor's. He lets "φν denote any sequence of rationals in [0, 1]." Cantor lets φν denote a sequence enumerating the rationals in [0, 1], which is the kind of sequence needed for his construction of a bijection between [0, 1] and the irrationals in (0, 1).Cantor's method can be used to modify the function {{nowrap|f{{space|hair}}2(t) {{=}} 0.t2}} to produce a bijection from T to (0, 1). Because some numbers have two binary expansions, {{nowrap|f{{space|hair}}2(t)}} is not even injective. For example, {{nowrap|f{{space|hair}}2(1000...) {{=}}}} 0.1000...2 = 1/2 and {{nowrap|f{{space|hair}}2(0111...) {{=}}}} 0.0111...2 = {{nowrap|1/4 + 1/8 + 1/16 + ... {{=}}}} 1/2, so both 1000... and 0111... map to the same number, 1/2.To modify {{nowrap|f2{{space|hair}}(t)}}, observe that it is a bijection except for a countably infinite subset of (0, 1) and a countably infinite subset of T. It is not a bijection for the numbers in (0, 1) that have two binary expansions. These are called dyadic numbers and have the form {{nowrap|m{{space|hair}}/{{space|hair}}2n}} where m is an odd integer and n is a natural number. Put these numbers in the sequence: r = (1/2, 1/4, 3/4, 1/8, 3/8, 5/8, 7/8, ...). Also, {{nowrap|f2{{space|hair}}(t)}} is not a bijection to (0, 1) for the strings in T appearing after the binary point in the binary expansions of 0, 1, and the numbers in sequence r. Put these eventually-constant strings in the sequence: s = ({{color|#808080|000}}..., {{color|#808080|111}}..., 1{{color|#808080|000}}..., 0{{color|#808080|111}}..., 01{{color|#808080|000}}..., 00{{color|#808080|111}}..., 11{{color|#808080|000}}..., 10{{color|#808080|111}}..., ...). Define the bijection g(t) from T to (0, 1): If t is the nth string in sequence s, let g(t) be the nth number in sequence r{{space|hair}}; otherwise, g(t) = 0.t2.To construct a bijection from T to R, start with the tangent function tan(x), which is a bijection from (−π/2, Ï€/2) to R (see the figure shown on the right). Next observe that the linear function h(x) = {{nowrap|Ï€x – Ï€/2}} is a bijection from (0, 1) to (−π/2, Ï€/2) (see the figure shown on the left). The composite function tan(h(x)) = {{nowrap|tan(Ï€x – Ï€/2)}} is a bijection from (0, 1) to R. Composing this function with g(t) produces the function tan(h(g(t))) = {{nowrap|tan(Ï€g(t) – Ï€/2)}}, which is a bijection from T to R.

General sets

File:Diagonal argument powerset svg.svg|thumb|250px|Illustration of the generalized diagonal argument: The set T = {n∈mathbb{N}: n∉f(n)} at the bottom cannot occur anywhere in the range of f:mathbb{N}→P(mathbb{N}). The example mapping f happens to correspond to the example enumeration s in the above picture.]]A generalized form of the diagonal argument was used by Cantor to prove Cantor's theorem: for every set S, the power set of S—that is, the set of all subsets of S (here written as P(S))—cannot be in bijection with S itself. This proof proceeds as follows:Let f be any function from S to P(S). It suffices to prove f cannot be surjective. That means that some member T of P(S), i.e. some subset of S, is not in the image of f. As a candidate consider the set: For every s in S, either s is in T or not. If s is in T, then by definition of T, s is not in f(s), so T is not equal to f(s). On the other hand, if s is not in T, then by definition of T, s is in f(s), so again T is not equal to f(s); cf. picture.For a more complete account of this proof, see Cantor's theorem.

Consequences

Ordering of cardinals

With equality defined as the existence of a bijection between their underlying sets, Cantor also defines binary predicate of cardinalities |S| and |T| in terms of the existence of injections between S and T. It has the properties of a preorder and is here written "le". One can embed the naturals into the binary sequences, thus proving various injection existence statements explicitly, so that in this sense |{mathbb N}|le|2^{mathbb N}|, where 2^{mathbb N} denotes the function space {mathbb N}to{0,1}. But following from the argument in the previous sections, there is no surjection and so also no bijection, i.e. the set is uncountable. For this one may write |{mathbb N}|