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I am familiar with notation like

  • $\mathbb{R}^x$ (meaning the set of all x-tuples with real elements)
  • $\mathbb{A}\times \mathbb{B}$ (being the set of all pairs whose first element is from $\mathbb{A}$ and second element from $\mathbb{B}$.

But what does the notation $\mathbb{A}^{\mathbb{R}\times\mathbb{R}}$ mean?

Ubiquitous
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  • @DietrichBurde I use $\mathbb{A}$ as a placeholder for an arbitrary set. But, for my purposes, it can be taken to be some subset of $\mathbb{R}$. – Ubiquitous Mar 28 '17 at 12:49

2 Answers2

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Usually, the notation $A^B$ denotes the set of all functions from $B$ to $A$.

This makes sense because in a sense, $\mathbb R^3$ can be seen as the set of all functions from $\{1,2,3\}$ to $\mathbb R$. This is because

  1. any function $f$ from $\{1,2,3\}$ to $\mathbb R$ can represent one element of $\mathbb R^3$ as $[f(1), f(2), f(3)]$, and
  2. every element $[x_1,x_2,x_3]$ of $\mathbb R^3$ represents one function from $\{1,2,3\}$ to $\mathbb R$, specifically the function for which $f(1)=x_1, f(2)=x_2$ and $f(3)=x_3$.

This generalizes even if we replace $3$ with an infinite set. For example, $\mathbb R^\mathbb N$ can be seen as the set of sequences of real numbers, so an element would be $[x_1,x_2,x_3,\dots]$, but at the same time, this is also a mapping from $\mathbb N$ to $\mathbb R$ (one that maps $1$ to $x_1$, $2$ to $x_2$ and so on).

In general then, $A^B$ simply denotes a set where each element is a "$|B|$-tuple", i.e. each elements is some mapping that maps each element of $B$ to some element of $A$. A function from $\mathbb R $ to $\mathbb R$ is really nothing more than an object which, for each real number $x$, perscribes another real number $y$. It's just our decision to denote this as $f(x)=y$.

5xum
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The set of all functions from $\mathbb{R}\times \mathbb{R}$ into $\mathbb{A}$. In general, $A^B$ usually refers to the set of functions from $B$ into $A$.

In fact, the notation $\mathbb{R}^n$ matches this; if we consider some $n$-element set, like $\{0,\ldots, n-1\}$, then an $n$-vector is exactly a map $\{0,\ldots, n-1\} \to \mathbb{R}$.

Hayden
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