The set $B^A$ of all functions $A\to B$

Let $A$ and $B$ be sets. By For sets $A$ and $B$ the collection of all functions $A\to B$ is a set, being a subset of $\mathcal{P}(A\times B)$ the functions $A\to B$ (A function is a relation $f$ with $(a,b)\in f$ and $(a,c)\in f$ implying $b=c$; $f:A\to B$, the value $f(a)$, domain and codomain) form a set; it is written

$B^A:=\{f:f\text{ is a function with }\mathrm{dom}f=A\text{ and }\mathrm{ran}f\subseteq B\}\mathrm{.}$

Thus $f\in B^A$ holds if and only if $f:A\to B$.

• The notation collides with exponentiation, and the collision is deliberate. $B^A$ is the same symbol the library already uses for a space of functions in The vector space $F^X$ of all functions $X\to F$ with pointwise operations, and $F^n$ as the case $X=n=\{0,1,\dots ,n-1\}$ ↗, where $F^X$ is the set of functions $X\to F$ carrying a vector-space structure; the underlying set there is the one defined here. Some set-theory texts write ${}^{A}B$ instead, reserving $B^A$ for ordinal and cardinal exponentiation. This library keeps $B^A$ for the function set, and Ordinal ${\alpha }^{\beta }$ and cardinal ${\kappa }^{\lambda }$ are different operations that share one notation ↗ is where the arithmetic exponentiations are distinguished from one another.

• Degenerate cases. $B^{\mathrm{\varnothing }}$ has exactly one element, the empty function, for every $B$, since a function with empty domain has no elements at all. ${\mathrm{\varnothing }}^A$ is empty whenever $A$ is nonempty, since a function with domain $A$ must take a value at each point of $A$, and ${\mathrm{\varnothing }}^{\mathrm{\varnothing }}$ again has the empty function as its only element.