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{{math proof| <math> B=\{x \in A \mid x \notin f(x)\}</math> exists via the [[axiom schema 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> From ''for all'' <math>x</math> ''in'' <math>A \ [ x \in B \iff x \notin f(x) ]</math> , we deduce <math>\xi \in B \iff \xi \notin f(\xi) </math> via [[universal instantiation]]. <br> The previous deduction yields a contradiction of the form <math>\varphi \Leftrightarrow \lnot \varphi</math>, since <math>f(\xi)=B</math>. <br> Therefore, <math>f</math> is not surjective, via [[reductio ad absurdum]]. <br> We know [[injective function|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> such that <math>g(x) = \{x\}</math>. <br> Consequently, <math>\operatorname{card}(A) < \operatorname{card}(\mathcal{P}(A))</math>. ∎}}
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> {{nowrap|to <math>Y</math>.}}
:<math>B=\{x\in A \mid x\not\in f(x)\}.</math>
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