:<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'' &epsilonisin; ''B'' or not. If ''y'' &epsilonisin; ''B'', then ''y''&epsilonisin; ''f''(''y''), but that implies, by definition of ''B'', that ''y'' '''not''' &epsilonnotin; ''B''. On the other hand if it is ''y'' '''not''' &epsilonnotin; ''B'', then ''y'' '''not''' &epsilonnotin; ''f''(''y'') and therefore ''y''&epsilonisin; ''B''. Either way, we get a contradiction.
Because of the double occurrence of ''x'' in the expression "''x'' not &isinnotin; ''x''", this is a [[diagonal argument]].
