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:<math>B=\left\{\,x\in A : x\not\in f(x)\,\right\}.</math>
To show that ''B'' is not in the image of ''f'', suppose that ''B'' ''is'' in the image of ''f''. Then for some ''y'' in ''A'', we have ''f''(''y'') = ''B''. Now consider whether ''y'' is a member of ''B'' or not. If ''y'' ε ''B'', then ''y'' ε ''f''(''y''), but that implies, by definition of ''B'', that ''y'' is '''not''' ε ''B''. On the other hand if it is '''not''' the case that ''y'' ε ''B'', then it is '''not''' true that ''y'' ε ''f''(''y'') and therefore it '''is''' true that ''y'' ε ''B''. Either way, we get a contradiction.
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