Let $S$ be a set. What does $S^z$ mean for each $z\in\mathbb{C}$?
In Set Theory numbers are sets and for any two sets $A$ and $B$, we define $B^A$ as the set of maps from $A$ to $B$. Well okay, but I don't understand what $S^z$ would mean (if anything) for $z\in\mathbb{C}\setminus\mathbb{N}$, where $0\notin\mathbb{N}$.
My Attempt.
I suppose $S^{0}=\emptyset$ would be a sensible guess.
Let $n\in\mathbb{N}$. Through Category Theory, in $\mathbf{\text{Set}}$ we can see the Cartesian product of $n$ copies of $S$ via cones in terms of commutative diagrams like so: the product is the cone $$\left(\prod_{r=1}^{n}{S}\stackrel{\pi_{i}}{\to} S\right)_{i=1}^n$$ such that for any cone $$\left(X\stackrel{p_{i}}{\to} S\right)_{i=1}^n$$ there exists a unique $u:X\to\prod_{r=1}^{n}{S}$ such that the following diagrams commute:
.
If $z\in \mathbb{R}$ and $z>0$, I suppose we could let $n$ be the integer part $[z]$ of $z$ in the above and conjure some appropriate $\mathbf{\text{Set}}$-arrow $\pi_{\{z\}}:\prod_{z}{S}\to \bar{S}$ for the fractional part $\{z\}$ of $z$ to get something like
.
I have no idea what $\pi_{\{z\}}:\prod_{z}{S}\to \bar{S}$ should be though.
To illustrate this sketch, if I could order the elements of $S$ with some $<$ to get $S_<$, I suppose I could take the first $\{z\}$ elements of $S_<$. For example, I'd have something like $\{1, 2, 3, 4\}^{2.5}:=\{1, 2, 3, 4\}^2\times\{1, 2\}$ under the natural order.
If the above makes sense, I'd try extending this to $z\in\mathbb{R}$ & $z<0$ using coproducts like so:
.
(I hope you'll forgive me for not defining things in the diagram immediately above.)
I need $S^{-z}\times S^{z}=S^0$.
If we return to Set Theory, I suppose we'd need to specify which construction of $\mathbb{R}$ we're using first to make sense of $S^z$ for $z\in\mathbb{R}$.
I have no idea what to make of $z\in\mathbb{C}\setminus\mathbb{R}$.
Is any of this making sense?
