Non-natural dimension of sets

Question

This question is not for an exercise or a task, but just mathematical curiosity:
In Mathematics we are used to work with sets such as $\mathbb{Z}$, $\mathbb{Q}$, etc. However, we also use their "powers" provided by the cartesian product ($\mathbb{R}^n$ for example). However, I wonder if somebody knows whether it is possible to define "non-natural" "powers" of this sets, or of any set apart from the usual ones, the same as with the set of real numbers, where exponentiation can be extended to any number like for example, $2^\pi$ or $3^{5.43}$. Is it possible to define stuff like $\mathbb{R}^{\frac{2}{5}}$, $\mathbb{Q}^\pi$ or $[0,1]^{\frac{\sqrt{2}}{3}}$? Does it makes sense? Has anybody thought about it before? Not only that, but if they exist, is it possible to equip these sets with an operation to create an algebraic structure? As a lover and still student of algebraic structures, it would be great for me that groups like $(\mathbb{R}^{\pi},+)$ could be created, or even more.
Postdata: I don't know which tags use for this question, so please don't downvote just for this.

Answer 1

For a set $X$ we may define $\sqrt{X}$ as a set (if it exists) with $\sqrt{X}^2 \cong X$. Here, recall that two sets are isomorphic when there is a bijection (=isomorphism) between them, and $Y^2$ is the cartesian product of $Y$ with itself.

Some easy observations:

• If $X$ is finite with $n$ elements, then $\sqrt{X}$ exists iff $n$ is a square number.
• If $X$ is infinite, then $\sqrt{X}$ exists, and in fact $\sqrt{X} \cong X$. This is because for every infinite set $X$ we have $X^2 \cong X$.
• If $\sqrt{X}$ exists, then it is uniquely determined up to isomorphism.

The main obstacle, though, is that there is no general, natural construction of $\sqrt{X}$ (by this I mean a functor from the category of sets which have a root to the category of sets). I think that this makes this notion pretty much useless. In particular, when thinking of $Y^2$ as the set of maps from a $2$-element set to $Y$, we cannot define $\sqrt{X}$ as set of maps from a $\frac{1}{2}$-element set (what should that be?) to $X$, at least not classically*.

A similar definition works in every cartesian cateory. For example, if $G$ is a group, we can define $\sqrt{G}$ as a group with $\sqrt{G}^2 \cong G$ (isomorphism of groups), if it exists. This propery is much more restrictive when compared to the category of sets. Also, $\sqrt{G}$ will not be uniquely determined anymore by MO/150793 (it is in the case of finite groups, though).

Similar remarks apply to the notion of $\sqrt[n]{X}$. For example, MO/10128 has examples of abelian groups with $A \cong \sqrt[3]{A} \not\cong \sqrt{A}$. The (must-read!) paper Seven Trees in One by Andreas Blass shows that the set of finite binary trees $T$ satisfies $T \cong \sqrt[7]{T}$, or more precisely that you can write down an isomorphism $T^7 \cong T$ explicitly instead of just bailing out with the argument that $T$ is infinite.

In ring theory, the square root notation is already taken for something else, though: If $I$ is an ideal of a ring $R$, then $\sqrt{I}$ is the radical of $I$. This operation is quite different from the square roots of numbers, though. For example, $\sqrt{\sqrt{I}}=\sqrt{I}$, and in $R=\mathbb{Z}$ we have $\sqrt{n}=\langle p_1 \cdots p_s \rangle$ when $p_1,\dotsc,p_s$ are the distinct primes dividing $n$.

*Janelidze and Street have investigated real sets (Tbilisi Mathematical Journal 10(3) (2017) 23-49) which presumably (I didn't read it in detail yet) enables us to find a "set" $S$ with $\frac{1}{2}$ elements. Then the "set" of maps $S \to X$ (if that exists in their framework) could be a version of $\sqrt{X}$, and $S \times X$ could be a version of $\frac{1}{2} X$.