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Please help with this question. Show that for a finite set $A$ of cardinality $n$, the cardinality of P(A) is $2^n$, where $P(A)$ is the power set of $A$. Thank you in advance for any help that is given.

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I changed $2n$ to $2^n$, and edited in some TeX. – Gerry Myerson Sep 6 '11 at 2:33
possible duplicate of Proving number of subsets of a set size $n$ via induction – Asaf Karagila Sep 6 '11 at 6:16
up vote 5 down vote accepted

Think of how you could go about constructing a subset of $A$. For each element, you choose to either include it or exclude it from the subset you are building. That gives you 2 choices for each of the $n$ elements of $A$. Multiplying your choices together, you get $2^n$ total possibilities. That is, there are $2^n$ different subsets you can build from $A$.

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Suppose I am choosing elements to put in some subset of the power set. Then each element can either be in, or not be in, my subset. So this means that altogether...

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Well done. A good hint. – Ross Millikan Sep 6 '11 at 4:23

Number the elements of $A$ as $a_1, a_2, \dots, a_n$. Consider the map $\chi:P(A) \to \{0,1\}^n$ given by $\chi(X)=(x_1,x_2,\dots,x_n)$, where $x_i=1$ iff $a_i \in X$. Then $\chi$ is a bijection.

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This seems like rather sophisticated language and notation for such an elementary question. Is there really anything here that is not in, say, Austin's answer? – Pete L. Clark Sep 6 '11 at 4:14

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