|A^A|<=|2^(2^A)|
in [Math]
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From: ksoileau on 22 Mar 2010 16:24 Does anyone have a proof that for any infinite set A, |A^A|<=| 2^(2^A)|? Thanks for any insight, Kerry M. Soileau
From: Arturo Magidin on 22 Mar 2010 17:26 On Mar 22, 3:24 pm, ksoileau <kmsoil...(a)gmail.com> wrote: > Does anyone have a proof that for any infinite set A, |A^A|<=| > 2^(2^A)|? In fact, |A^A| = |2^A| and |A^A| < |2^(2^A)|. Trivially, |2^A| <= |A^A|, since any function from A to 2 can be realized as a function from A to A. For the other inequality, you have |A|<|2^A|, so |A^A| <= |(2^A)^A| = |2^A|^|A| = |2^(|A||A|)| = |2^|A| | = |2^A| (since |A||A|=|A|, as |A| is infinite). Since |A^A| = |2^A|, and |X| < |2^X| for all sets X (finite or otherwise), then for any infinite set A you have |A^A| < |2^(A^A)| = | 2^(2^A)|. -- Arturo Magidin
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