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What's the meaning of a set to the power of another set?

What does $\mathbb R^S$ mean when $S$ is a set? I am reading a text and I wonder if it has a special meaning or it's just a typo? Note that $\mathbb R^{|S|}$, where $|S|$ is the size of $S$, makes sense in the context.

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marked as duplicate by Pedro Tamaroff, Gerry Myerson, anon, sdcvvc, Ayman Hourieh Sep 30 '12 at 18:00

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As anon points out, it is just notation for the set of all functions $S \to \mathbb{R}$. When I first saw this notation, I found it confusing, but one of the benefits is that

$$\left|B^A\right| = |B|^{|A|}$$

where $|\cdot|$ denotes cardinality. You have to use cardinal arithmetic for infinite sets, but it still works.

A special case of this relationship is the fact that $|\mathcal{P}(A)| = 2^{|A|}$. To see this, first note that there is a $1-1$ correspondence between subsets of $A$ and functions $A \to \{0, 1\}$ given by the indicator function ($1$ if it is in the subset, $0$ if not). Then we have:

$$|\mathcal{P}(A)| = \left|\{0, 1\}^A\right| = |\{0, 1\}|^{|A|} = 2^{|A|}.$$

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I would interpret it to mean the set of functions from $S$ to ${\Bbb R}$.

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Also, if $[n]$ denotes the set $\{1,2,\dots,n\}$, then we can canonically identify $\mathbb{R}^n$ with $\mathbb{R}^{[n]}$, so the notation is consistent with the more familiar notation. –  Michael Joyce Sep 30 '12 at 8:00
    
Hmm, I always thought of $n$ as being synonymous with $\{0,1,\ldots,n-1\}$, which means that ${\Bbb R}^n$ is just ${\Bbb R}^n$. –  user22805 Sep 30 '12 at 17:50

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