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Suppose $A = \{ 1,...,n \}$. Prove that $\exists$ a bijection of $\mathcal{P}(A) = \{ B : B \subset A \}$ (power set of A) and the set $X^n $ where $X = \{ 0,1 \}$. In addition, Prove that $\mathcal{P}(A)$ if finite.

my work:

I have an idea how to construct this bijection: Notice that ${n \choose i}$ is the number of subsets of $A$ of size $i$. Now, let $B \subseteq A$. Let $N(B)$ denote the sum of the elements of $B$. For example, if $ B = \{1,2,3 \} \subset A$, then $N(B) = 1+2+3 = 6$ and we can define $f : \mathcal{P}(A) \to X^n$ as

$$ f(B) = (\underbrace{1,1,1,....,1}_{N(B) times}, 0 , 0,0,0,0 ) $$

This is still not a bijection as elements of the form (1,1,0,1,1,1,....) have no $B$ that get mapped to it. Any hint into how to construct this function? Is my strategy correct?

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    $\begingroup$ Maybe take a small, fixed value for $n$ (e.g. 3) and explicitly write out its power set, then see if there's a nice way to associate it to triplets of 1s and 0s? I would suggest making the empty set map to (0, 0, 0) and {1, 2, 3} map to (1, 1, 1), if that's a bit of a hint. $\endgroup$ – ConMan May 13 '20 at 2:22
  • $\begingroup$ I don't think your function is well defined,. Any $B \subseteq A$ that contains $n$ and any other element would have $N(B) >n$, but you only have $n$ many spaces for 0s and 1s. Also, it isn't injective as $\{1,2\}$ and $\{3\}$ have the same value under $B$. The bijection you want doesn't really care that $A$ contains exactly the numbers $1,...,n$. It only cares that it contains exactly $n$ elements, so you shouldn't be looking for a "numerical" function. $\endgroup$ – James May 13 '20 at 2:49
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The natural way to do this is to define $f_S\in X^n$ by $f_S(x)=\begin{cases}1\,,x\in S\\0,\,x\notin S\end{cases}$.

This defines a bijection between $P(A)$ and the function space $X^n$.

You can see from this that $P(A)$ has order $2^n$.

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