If $A$ is a any set and $B = \{0,1\}$, how do I show $P(A) \sim Map(A,B)$.

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Can someone provide me a detailed instruction? I heard this place is really helpful.

Also, you asked the same question earlier today: http://math.stackexchange.com/questions/331661/if-a-is-a-set-then-mathrmcardpa-2-mathrmcard-a — Asaf Karagila, Mar 16 '13 at 18:15

score 5 · Answer 1 · answered Mar 16 '13 at 17:48

Let $X$ be a subset of $A$ and define the function $\chi_X:A\to \{0,1\}$ by $$\chi_X(a)=\begin{cases} 1\; ;\text{ if } a\in X\\0\; ;\text{ if }a\in A\setminus X\end{cases}$$

This function simply says "yes" ($1$) or "no" ($0$) to the question "Is $a\in X$?"

Consider the map $\mu:P(A)\to B^A$ defined by $\mu(X)=\chi_X$. If $\chi_X$ is identical to $\chi_Y$ it follows by definition that $X=Y$, so this is injective. On the other hand, every $f\in B^A$ defines a subset of $A$ by $X_f=\{x\in A:f(x)=1\}$. So the function is a bijection.

score 2 · Answer 2 · answered Mar 16 '13 at 17:48

$P(A)$ is the set of all subsets of $A$. Choosing a subset $S$ of $A$ consists exactly in choosing for each element of $A$ whether it belongs to $S$ or not, which can be seen as assigning $1$ to $a \in A$ if $a \in S$ and $0$ otherwise, which is exactly defining a map from $A$ to $B$.

If $A$ is a any set and $B = \{0,1\}$, how do I show $P(A) \sim Map(A,B)$.

2 Answers2