Cantor's theorem: Difference between revisions

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{{math proof| Consider the set <math> B=\{x \in A \mid x \notin f(x)\}</math>. Suppose to the contrary that <math>f</math> is surjective. Then there exists <math>\xi\in A</math> such that <math>f(\xi)=B</math>. But by construction, <math>\xi \in B \iff \xi \notin f(\xi)= B </math>. This is a contradiction. Thus, <math>f</math> cannot be surjective. On the other hand, <math>g : A \to \mathcal{P}(A)</math> defined by <math>x \mapsto \{x\}</math> is an injective map. 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|<math>Y</math>.}} It suffices to show that there is no surjection from <math>X</math> 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> ∈ <math>A</math>. (Recall that each <math>f(x)</math> is a subset of <math>A</math>.) Such 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>