Does it make sense to try to extend the concept of a raised set to a real number? The definition of $a^b$ when $a$ and $b$ are natural numbers involves repeated multiplication. This definition can be extended to arbitrary exponents, motivated by the desire to preserve the identity $a^x\times a^y=a^{x+y}$.
Is it possible to extend the definition of the cartesian product in an analogous manner? It is clear what $\mathbb{R}^2$ or $\mathbb{R}^3$ means, but it is not obvious how $\mathbb R^{t}$ should be defined for arbitrary real values of $t$.
 A: Unfortunately there really isn't anything like that. This is a reasonable question to ask.  We can generalize ordinary exponentiation to work on sets instead of numbers, and all the familiar properties still hold, if you interpret them correctly.  For example, we can say that $\Bbb R^2 \times\Bbb R^1 = \Bbb R^3$ and have it mean something interesting and useful.
To make it work we have to interpret “$2$” and “$3$” as sets with $2$ and $3$ elements.  But this doesn't work for $\Bbb R^{1.5}$ because the idea of a set with $1.5$ elements seems meaningless.  Perhaps we could make sense of it somehow?  This has been investigated and the answer is: probably not.  If we did make sense of it, then whatever $\Bbb R^{1.5}$ should mean, it ought to at least have the property that $$\Bbb R^{1.5} \times \Bbb R^{1.5} = \Bbb R^3.$$
Could we find some mathematical object that behaves like this? This question has been addressed in the mathematical literature, and the answer is known:  there is no such object.

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*"Another Proof That $\Bbb R^3$ Has No Square Root", Sam B. Nadler, Jr., American Mathematical Monthly vol 111 June–July 2004, pp. 527–528.


*"$\Bbb R^3$ Has No Root", Robbert Fokkink, American Mathematical Monthly vol 109 March 2002, p. 285
You can make anything work if you are willing to ignore enough of the ways in which it doesn't work. But the only way to make this work seems to be to disregard all the interesting special structure of $\Bbb R^3$. For example, if you were to interpret “$=$” so that  $\Bbb R^3 = \Bbb R^2 = \Bbb R$, you could perhaps make $\Bbb R^{1.5}$ work.  I think this isn't what you want though.
(I like the Fokkink paper better, it's simpler.)
