Cantor's theorem: Difference between revisions

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In mathematical [[set theory]], '''Cantor's theorem''' is a fundamental result which states that, for any [[Set (mathematics)|set]] <math>A</math>, the set of all [[subset]]s of <math>A,</math> known as the [[power set]] of <math>A,</math> has a strictly greater [[cardinality]] than <math>A</math> itself.

For [[finite set]]s, Cantor's theorem can be seen to be true by simple [[enumeration]] of the number of subsets. Counting the [[empty set]] as a subset, a set with <math>n</math> elements has a total of <math>2^n</math> subsets, and the theorem holds because <math>2^n > n</math> for all [[non-negative integers]].

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Much more significant is Cantor's discovery of an argument that is applicable to any set, and shows that the theorem holds for [[infinite set|infinite]] sets also. As a consequence, the cardinality of the [[real number]]s, which is the same as that of the power set of the [[integer]]s, is strictly larger than the cardinality of the integers; see [[Cardinality of the continuum]] for details.

The theorem is named for German [[mathematician]] [[Georg Cantor]], who first stated and proved it at the end of the 19th century. Cantor's theorem had immediate and important consequences for the [[philosophy of mathematics]]. For instance, by iteratively taking the power set of an infinite set and applying Cantor's theorem, we obtain an endless hierarchy of infinite cardinals, each strictly larger than the one before it. Consequently, the theorem implies that there is no largest [[cardinal number]] (colloquially, "there's no largest infinity").

==Proof==

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{{math theorem|name=Theorem (Cantor)|math_statement=Let <math>f</math> be a map from set <math>A</math> to its power set <math>\mathcal{P}(A)</math>. Then <math>f : A \to \mathcal{P}(A)</math> is not [[surjective]]. As a consequence, <math>\operatorname{card}(A) < \operatorname{card}(\mathcal{P}(A))</math> holds for any set <math>A</math>.}}

{{math proof| Consider the set <math> B=\{x \in A \mid x \notin f(x)\}</math>. exists Suppose tovia the contrary[[axiom thatschema of specification]], and <math> B \in \mathcal{P}(A) </math> because <math> B \subseteq A </math>. <br> Assume <math> f </math> is surjective. <br> Then there exists a <math>\xi \in A</math> such that <math>f(\xi)=B</math>. <br> ButFrom by&nbsp;''for all'' <math>x</math> ''in'' <math>A \ [ x \in B \iff x \notin f(x) ]</math> construction, we deduce &nbsp;<math>\xi \in B \iff \xi \notin f(\xi)= B </math> &nbsp;via [[universal instantiation]]. <br> ThisThe isprevious deduction yields a contradiction of the form <math>\varphi \Leftrightarrow \lnot \varphi</math>, since <math>f(\xi)=B</math>. <br> ThusTherefore, <math>f</math> cannotis benot surjective, via [[reductio ad absurdum]]. <br> OnWe theknow other[[injective handfunction|injective maps]] from <math>A</math> to <math>\mathcal{P}(A)</math> exist. For example, a function <math>g : A \to \mathcal{P}(A)</math> definedsuch bythat <math>g(x) \mapsto= \{x\}</math> is an injective map. <br> Consequently, we must have <math>\operatorname{card}(A) < \operatorname{card}(\mathcal{P}(A))</math>. [[Q.E.D.]]∎}}

By definition of cardinality, we have <math>\operatorname{card}(X) < \operatorname{card}(Y)</math> for any two sets <math>X</math> and <math>Y</math> if and only if there is an [[injective function]] but no [[Bijective Function|bijective function]] from <math>X</math> to {{nowrap|to <math>Y</math>.}} It suffices to show that there is no surjection from <math>X</math> {{nowrap|to <math>Y</math>}}. This is the heart of Cantor's theorem: there is no surjective function from any set <math>A</math> to its power set. To establish this, it is enough to show that no function <math>f</math> (that maps elements in <math>A</math> to subsets of <math>A</math>) can reach every possible subset, i.e., we just need to demonstrate the existence of a subset of <math>A</math> that is not equal to <math>f(x)</math> for any <math>x</math> ∈\in <math>A</math>. (RecallRecalling that each <math>f(x)</math> is a subset of <math>A</math>.), Suchsuch a subset is given by the following construction, sometimes called the [[Cantor's diagonal argument|Cantor diagonal set]] of <math>f</math>:<ref name="Dasgupta2013">{{cite book|author=Abhijit Dasgupta|title=Set Theory: With an Introduction to Real Point Sets|year=2013|publisher=[[Springer Science & Business Media]]|isbn=978-1-4614-8854-5|pages=362–363}}</ref><ref name="Paulson1992">{{cite book|author=Lawrence Paulson|title=Set Theory as a Computational Logic |url=https://www.cl.cam.ac.uk/techreports/UCAM-CL-TR-271.pdf|year=1992|publisher=University of Cambridge Computer Laboratory|page=14}}</ref>

:<math>B=\{x\in A \mid x\not\in f(x)\}.</math>

This means, by definition, that for all ''<math>x''&nbsp;∈&nbsp;''\in A''</math>, ''<math>x''&nbsp;∈&nbsp;''\in B''</math> if and only if ''<math>x''&nbsp;∉&nbsp;''\notin f''(''x'')</math>. For all ''<math>x''</math> the sets ''<math>B''</math> and ''<math>f''(''x'')</math> cannot be the sameequal because ''<math>B''</math> was constructed from elements of ''<math>A''</math> whose [[Image (mathematics)|images]] (under ''<math>f'')</math> did not include themselves. MoreFor specifically, consider anyall ''<math>x''&nbsp;∈&nbsp;''\in A'', then</math> either ''<math>x''&nbsp;∈&nbsp;''\in f''(''x'')</math> or ''<math>x''&nbsp;∉&nbsp;''\notin f''(''x'')</math>. InIf the<math>x\in formerf(x)</math> case,then ''<math>f''(''x'')</math> cannot equal ''<math>B''</math> because ''<math>x''&nbsp;∈&nbsp;''\in f''(''x'')</math> by assumption and ''<math>x''&nbsp;∉&nbsp;''\notin B''</math> by the construction of ''B''definition. InIf the<math>x\notin latterf(x)</math> case,then ''<math>f''(''x'')</math> cannot equal ''<math>B''</math> because ''<math>x''&nbsp;∉&nbsp;''\notin f''(''x'')</math> by assumption and ''<math>x''&nbsp;∈&nbsp;''\in B''</math> by the constructiondefinition of ''<math>B''</math>.

Equivalently, and slightly more formally, we have just proved that the existence of ξ<math>\xi ∈\in ''A''</math> such that ''<math>f''(ξ)\xi )= ''B''</math> implies the following [[contradiction]]:

Equivalently, and slightly more formally, we just proved that the existence of ξ ∈ ''A'' such that ''f''(ξ) = ''B'' implies the following [[contradiction]]:

:<math>\begin{aligned}

\xi\notinin f(\xi)B &\iff \xi\innotin Bf(\xi) && \text{(by definition of }B\text{)}; \\

\xi \in B &\iff \xi \in f(\xi) && \text{(by assumption that }f(\xi)=B\text{)};. \\

\end{aligned}</math>

Therefore, by [[reductio ad absurdum]], the assumption must be false.<ref name="Priest2002"/> Thus there is no ξ<math>\xi ∈\in ''A''</math> such that ''<math>f''(ξ)\xi )= ''B''</math> ; in other words, ''<math>B''</math> is not in the image of ''<math>f''</math> and ''<math>f''</math> does not map toonto every element of the power set of ''<math>A''</math>, i.e., ''<math>f''</math> is not surjective.

Finally, to complete the proof, we need to exhibit an injective function from ''<math>A''</math> to its power set. Finding such a function is trivial: just map ''<math>x''</math> to the singleton set <math>\{''x''\}</math>. The argument is now complete, and we have established the strict inequality for any set ''<math>A''</math> that <math>\operatorname{card}(''A'') < \operatorname{card}(𝒫\mathcal{P}(''A''))</math>.

Another way to think of the proof is that ''<math>B''</math>, empty or non-empty, is always in the power set of ''<math>A''</math>. For ''<math>f''</math> to be [[Surjective function|onto]], some element of ''<math>A''</math> must map to ''<math>B''</math>. But that leads to a contradiction: no element of ''<math>B''</math> can map to ''<math>B''</math> because that would contradict the criterion of membership in ''<math>B''</math>, thus the element mapping to ''<math>B''</math> must not be an element of ''<math>B''</math> meaning that it satisfies the criterion for membership in ''<math>B''</math>, another contradiction. So the assumption that an element of ''<math>A''</math> maps to ''<math>B''</math> must be false; and ''<math>f''</math> cannot be onto.

Because of the double occurrence of <math>x</math> in the expression "<math>x\in f(x)</math>", this is a [[Cantor's diagonal argument|diagonal argument]]. For a countable (or finite) set, the argument of the proof given above can be illustrated by constructing a table in which

Because of the double occurrence of ''x'' in the expression "''x'' ∉ ''f''(''x'')", this is a [[Cantor's diagonal argument|diagonal argument]]. For a countable (or finite) set, the argument of the proof given above can be illustrated by constructing a table in which each row is labelled by a unique ''x'' from ''A'' = {''x''₁, ''x''₂, ...}, in this order. ''A'' is assumed to admit a [[Total order|linear order]] so that such table can be constructed. Each column of the table is labelled by a unique ''y'' from the [[power set]] of ''A''; the columns are ordered by the argument to ''f'', i.e. the column labels are ''f''(''x''₁), ''f''(''x''₂), ..., in this order. The intersection of each row ''x'' and column ''y'' records a true/false bit whether ''x'' ∈ ''y''. Given the order chosen for the row and column labels, the main diagonal ''D'' of this table thus records whether ''x'' ∈ ''f''(''x'') for each ''x''&nbsp;∈&nbsp;''A''. The set ''B'' constructed in the previous paragraphs coincides with the row labels for the subset of entries on this main diagonal ''D'' where the table records that ''x'' ∈ ''f''(''x'') is false.<ref name="Priest2002">{{cite book|author=Graham Priest|title=Beyond the Limits of Thought|year=2002|publisher=Oxford University Press|isbn=978-0-19-925405-7|pages=118–119}}<!--note that the page numbers differ between the OUP and CUP editions of Priest's book!--></ref> Each column records the values of the [[indicator function]] of the set corresponding to the column. The indicator function of ''B'' coincides with the [[logical negation|logically negated]] (swap "true" and "false") entries of the main diagonal. Thus the indicator function of ''B'' does not agree with any column in at least one entry. Consequently, no column represents ''B''.

# each row is labelled by a unique <math>x</math> from <math>A=\{x_1 ,x_2 , \ldots \}</math>, in this order. <math>A</math> is assumed to admit a [[Total order|linear order]] so that such table can be constructed.

# each column of the table is labelled by a unique <math>y</math> from the [[power set]] of <math>A</math>; the columns are ordered by the argument to <math>f</math>, i.e. the column labels are <math>f(x_1),f(x_2)</math>, ..., in this order.

# the intersection of each row <math>x</math> and column <math>y</math> records a true/false bit whether <math>x\in y</math>.

Given the order chosen for the row and column labels, the main diagonal <math>D</math> of this table thus records whether <math>x\in f(x)</math> for each <math>x\in A</math>. One such table will be the following:

<math display="block">\begin{array}{cccccc}

& f(x_1) & f(x_2) & f(x_3) & f(x_4) & \cdots \\

\hline

x_1 & {\color{red} T} & T & F & T & \cdots \\

x_2 & T & {\color{red} F} & F & F & \cdots \\

x_3 & F & F & {\color{red} T} & T & \cdots \\

x_4 & F & T & T & {\color{red} T} & \cdots \\

\vdots & \vdots & \vdots & \vdots & \vdots & \ddots

\end{array}</math>

Because of the double occurrence of ''x'' in the expression "''x'' ∉ ''f''(''x'')", this is a [[Cantor's diagonal argument|diagonal argument]]. For a countable (or finite)The set, the argument of the proof given above can be illustrated by constructing a table in which each row is labelled by a unique ''x'' from ''A'' = {''x''<submath>1B</submath>, ''x''₂, ...}, in this order. ''A'' is assumed to admit a [[Total order|linear order]] so that such table can be constructed. Each column of the table is labelled by a unique ''y'' from the [[power set]] of ''A''; the columns are ordered by the argument to ''f'', i.e. the column labels are ''f''(''x''₁), ''f''(''x''₂), ..., in this order. The intersection of each row ''x'' and column ''y'' records a true/false bit whether ''x'' ∈ ''y''. Given the order chosen for the row and column labels, the main diagonal ''D'' of this table thus records whether ''x'' ∈ ''f''(''x'') for each ''x''&nbsp;∈&nbsp;''A''. The set ''B'' constructed in the previous paragraphs coincides with the row labels for the subset of entries on this main diagonal ''<math>D''</math> (which in above example, coloured red) where the table records that ''<math>x''\in ∈ ''f''(''x'')</math> is false.<ref name="Priest2002">{{cite book|author=Graham Priest|title=Beyond the Limits of Thought|year=2002|publisher=Oxford University Press|isbn=978-0-19-925405-7|pages=118–119}}<!--note that the page numbers differ between the OUP and CUP editions of Priest's book!--></ref> Each columnrow records the values of the [[indicator function]] of the set corresponding to the column. The indicator function of ''<math>B''</math> coincides with the [[logical negation|logically negated]] (swap "true" and "false") entries of the main diagonal. Thus the indicator function of ''<math>B''</math> does not agree with any column in at least one entry. Consequently, no column represents ''<math>B''</math>.

Despite the simplicity of the above proof, it is rather difficult for an [[automated theorem prover]] to produce it. The main difficulty lies in an automated discovery of the Cantor diagonal set. [[Lawrence Paulson]] noted in 1992 that [[Otter (theorem prover)|Otter]] could not do it, whereas [[Isabelle (proof assistant)|Isabelle]] could, albeit with a certain amount of direction in terms of tactics that might perhaps be considered cheating.<ref name="Paulson1992"/>

==When ''A'' is countably infinite==

Let us examine the proof for the specific case when <math>A</math> is [[countably infinite]]. [[Without loss of generality]], we may take {{<math|1=''>A'' = '''\mathbb{N'''} = \{{mset|1, 2, 3, …}}}\ldots\}</math>, the set of [[natural number]]s.

Suppose that '''<math>\mathbb{N'''}</math> is [[equinumerous]] with its [[power set]] 𝒫<math>\mathcal{P}('''\mathbb{N'''})</math>. Let us see a sample of what 𝒫<math>\mathcal{P}('''\mathbb{N'''})</math> looks like:

:<math>\mathcal{P}(\mathbb{N})=\{\varnothing,\{1, 2\}, \{1, 2, 3\}, \{4\}, \{1, 5\}, \{3, 4, 6\}, \{2, 4, 6,\dots\},\dots\}.</math>

𝒫Indeed, <math>\mathcal{P}('''\mathbb{N'''})</math> contains infinite subsets of '''<math>\mathbb{N'''}</math>, e.g. the set of all positive even numbers <math>\{2, 4, 6,...\ldots\},=\{2k:k\in as\mathbb{N}\}</math>, wellalong aswith the [[empty set]] <math>\varnothing</math>.

Now that we have an idea of what the elements of 𝒫<math>\mathcal{P}('''\mathbb{N'''})</math> look likeare, let us attempt to pair off each [[element (math)|element]] of '''<math>\mathbb{N'''}</math> with each element of 𝒫<math>\mathcal{P}('''\mathbb{N'''})</math> to show that these infinite sets are equinumerous. In other words, we will attempt to pair off each element of '''<math>\mathbb{N'''}</math> with an element from the infinite set 𝒫<math>\mathcal{P}('''\mathbb{N'''})</math>, so that no element from either infinite set remains unpaired. Such an attempt to pair elements would look like this:

:<math>\mathbb{N}\begin{Bmatrix} 1 & \longleftrightarrow & \{4, 5\}\\ 2 & \longleftrightarrow & \{1, 2, 3\} \\ 3 & \longleftrightarrow & \{4, 5, 6\} \\ 4 & \longleftrightarrow & \{1, 3, 5\} \\ \vdots & \vdots & \vdots \end{Bmatrix}\mathcal{P}(\mathbb{N}).</math>

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Given such a pairing, some natural numbers are paired with [[subset]]s that contain the very same number. For instance, in our example the number 2 is paired with the subset {1, 2, 3}, which contains 2 as a member. Let us call such numbers ''selfish''. Other natural numbers are paired with [[subset]]s that do not contain them. For instance, in our example the number 1 is paired with the subset {4, 5}, which does not contain the number 1. Call these numbers ''non-selfish''. Likewise, 3 and 4 are non-selfish.

Using this idea, let us build a special set of natural numbers. This set will provide the [[proof by contradiction|contradiction]] we seek. Let ''<math>B''</math> be the set of ''all'' non-selfish natural numbers. By definition, the [[power set]] 𝒫<math>\mathcal{P}('''\mathbb{N'''})</math> contains all sets of natural numbers, and so it contains this set ''<math>B''</math> as an element. If the mapping is bijective, ''<math>B''</math> must be paired off with some natural number, say ''<math>b''</math>. However, this causes a problem. If ''<math>b''</math> is in ''<math>B''</math>, then ''<math>b''</math> is selfish because it is in the corresponding set, which contradicts the definition of ''<math>B''</math>. If ''<math>b''</math> is not in ''<math>B''</math>, then it is non-selfish and it should instead be a member of ''<math>B''</math>. Therefore, no such element ''<math>b''</math> which maps to ''<math>B''</math> can exist.

Since there is no natural number which can be paired with ''<math>B''</math>, we have contradicted our original supposition, that there is a [[bijection]] between '''<math>\mathbb{N'''}</math> and 𝒫<math>\mathcal{P}('''\mathbb{N'''})</math>.

Note that the set ''<math>B''</math> may be empty. This would mean that every natural number ''<math>x''</math> maps to a subset of natural numbers that contains ''<math>x''</math>. Then, every number maps to a nonempty set and no number maps to the empty set. But the empty set is a member of 𝒫<math>\mathcal{P}('''\mathbb{N'''})</math>, so the mapping still does not cover 𝒫<math>\mathcal{P}('''\mathbb{N'''})</math>.

Through this [[proof by contradiction]] we have proven that the [[cardinality]] of '''<math>\mathbb{N'''}</math> and 𝒫<math>\mathcal{P}('''\mathbb{N'''})</math> cannot be equal. We also know that the [[cardinality]] of 𝒫<math>\mathcal{P}('''\mathbb{N'''})</math> cannot be less than the [[cardinality]] of '''<math>\mathbb{N'''}</math> because 𝒫<math>\mathcal{P}('''\mathbb{N'''})</math> contains all [[singleton (mathematics)|singleton]]s, by definition, and these singletons form a "copy" of '''<math>\mathbb{N'''}</math> inside of 𝒫<math>\mathcal{P}('''\mathbb{N'''})</math>. Therefore, only one possibility remains, and that is that the [[cardinality]] of 𝒫<math>\mathcal{P}('''\mathbb{N'''})</math> is strictly greater than the [[cardinality]] of '''<math>\mathbb{N'''}</math>, proving Cantor's theorem.

==Related paradoxes==

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==Generalizations==

Cantor[[Lawvere's fixed-point theorem]] hasprovides beenfor generalizeda broad generalization of Cantor's theorem to any [[Category (mathematics)|category]] with [[Product (category theory)|finite products]] in the following way:<ref name="LawvereSchanuel2009">{{cite book|author1=F. William Lawvere|author2=Stephen H. Schanuel|title=Conceptual Mathematics: A First Introduction to Categories|year=2009|publisher=Cambridge University Press|isbn=978-0-521-89485-2|at=Session 29|url-access=registration|url=https://archive.org/details/conceptualmathem00lawv}}</ref> supposelet that<math>\mathcal{C}</math> be such a category, and let <math>Y1</math> isbe ana terminal object in a<math>\mathcal{C}</math>. categorySuppose that <math>CY</math> withis terminalan object in <math>1\mathcal{C}</math>, and that there exists an endomorphism <math>\alpha : Y \to Y</math> that does not have any fixed points; that is, forthere everyis no morphism <math>y:1 \to Y</math>, that satisfies <math>\alpha \circ y \ne= y</math>. Then there is no object <math>T</math> of <math>\mathcal{C}</math> such that a morphism <math>f: T \times T \to Y</math> can parameterize all morphisms <math>T \to Y</math>. In other words, for every object <math>T</math> and every morphism <math>f : T \times T \to Y</math>, an attempt to write maps <math>T \to Y</math> as maps of the form <math>f(-,x) : T \times T \to Y</math> must leave out at least one map <math>T \to Y</math>.

==See also==