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 September 4th, 2008, 09:28 AM #1 Newbie   Joined: Sep 2008 Posts: 3 Thanks: 0 abstract alg Please help me figure out how to get started with this problem.. For every positive integer n, prove that a set with exactly n elements has exactly 2^n subsets (counting the empty set and entire set).
 September 4th, 2008, 10:24 AM #2 Senior Member   Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 Re: abstract alg Try induction. The case for n=1 is simple; for n>1, let the set S have n elements and $2^n$ subsets, and let $x\notin S;$ then for each subset $T\subseteq S,$ both $T$ and $T\cup\{x\}$ are subsets of the set $S\cup\{x\}$ which contains n+1 elements. Moreover, these subsets are all the possible subsets of $S\cup\{x\}$ (you will need to show this); since there are two for each $T$, and there are $2^n$ possible subsets $T\subseteq S$, there must be $2^{n+1}$ possible subsets of the set $S\cup\{x\}.$

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### for every positive integer n, prove that a set with exactly n elements has exactly 2n subsets

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