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:

enter image description here.

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

enter image description here.

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:

enter image description here.

(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?

  • 3
    $\begingroup$ So, I guess you want to generalize the "cartesian product" to non-integer powers. So, perhaps the first thing to do is define $S^{1/2}$ somehow. If $|S|=2$, does that mean $S^{1/2}$ should have $|S^{1/2}| = \sqrt{2}$ ?? $\endgroup$ – GEdgar Dec 6 '14 at 22:13
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    $\begingroup$ Actually, $S^0$ should have one element, not zero elements. $\endgroup$ – GEdgar Dec 6 '14 at 22:14
  • $\begingroup$ I'm not sure about that take on $S^0$, @GEdgar. Please explain it :) $\endgroup$ – Shaun Dec 6 '14 at 22:24
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    $\begingroup$ Frankly, no, none of this makes any sense. Not everything is begging to be generalised. $\endgroup$ – Zhen Lin Dec 6 '14 at 22:49
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    $\begingroup$ Related: math.stackexchange.com/questions/198514/… $\endgroup$ – Berci Dec 7 '14 at 0:06

You probably don't want to read this and this is not really an answer, but your idea of defining $S^z$ for complex $z$ just doesn't make any sense. You can draw as many diagrams as you wish, this doesn't matter. Even for $z \in \mathbb{Z}$ there is no way to define $S^z$. The only invertible object in the monoidal category of sets is the one-point set, and of course $1^z=1$ is not really a discovery.

  • 1
    $\begingroup$ "You probably don't want to read this" - You're wrong. This is much more beautiful than the messy scenario I had in mind. Thank you! :D $\endgroup$ – Shaun Dec 7 '14 at 9:31

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