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Let $\mathcal{P}(S)$ denote the power set of $S$. Why is $\mathcal{P}(S) = 2^S$?

I know that the power set of $S = \{x,y,z\}$ is

$$\mathcal{P}(S) = \{\{\}, \{x\}, \{y\}, \{z\}, \{x, y\}, \{x, z\}, \{y, z\}, \{x, y, z\}\},$$

so how does this relate to $\mathcal{P}(S) = 2^S$? I cannot understand why

$$\mathcal{P}(S) = \mathcal{P}(\{x,y,z\}) = 2^{\{x,y,z\}} = \{\{\}, \{x\}, \{y\}, \{z\}, \{x, y\}, \{x, z\}, \{y, z\}, \{x, y, z\}\},$$

I assume it is somehow related tot he fact that the set $\{x,y,z\}$ has 3 elements, and $\mathcal{P}(\{x,y,z\})$ is a set of $2^3 = 8$ sets.

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    It's just another notation to denote the power set. You could ask about the motivation for this notation, but would be a different matter. – 2017-01-08

1 Answers 1

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In set theory, the notation $X^Y$ doesn't mean the same thing it means in arithmetic; instead, it means "the set of all functions from $Y$ to $X$". If you say that the number "2" means the set containing $0$ and $1$ (another set-theory trick!), then functions from $Y$ to $\{0, 1\}$ correspond, in a 1-1 way, with subsets of $Y$: if $A$ is a subset of $Y$, we can define a function $$ f_A(y) = \begin{cases} 1 & y \in A \\ 0 & y \notin A\end{cases}. $$

And the "fun" thing is that if $X$ and $Y$ are finite, and $X$ has $n$ elements and $Y$ has $k$ elements, then the number of elements of $X^Y$ happens to be $n^k$ (where this is the arithmetic exponential).

Roughly, this is all one big notational pun.