# Uncountable applications of an operator?

To simplify matters, assume we have a commutative group $$(X,\cdot,1)$$ with uncountable $$X$$. For commutative groups, applications of elements $$x_i\in X$$ don’t care about order, and we can simply count the number of times each element of $$X$$ is applied. This means that if we don’t apply any element in $$X$$ twice, (e.g. $$a\cdot b \cdot c$$ but not $$a\cdot a\cdot c$$), then we can simply associate with each application of elements, a set $$Y\subseteq X$$ (This is just to simplify the question but not essential)

This means we can represent the $$\cdot:X\times X\to X$$ operator instead as a function $$\square : \mathcal P (X) \to X$$.

Normally, this set inputted in $$\square$$ will be finite or at least countable, because when we do applications of elements of $$X$$ we list this as a sequence of applications.

But the type signature of $$\square$$ suggests that we can also input uncountable subsets of $$X$$. Is there a theory about this? This would be an uncountable application of an operator.

• This is utter nonsense. You cannot define $\square$ on infinite sets at all, countable or uncountable. – Eric Wofsey Feb 10 at 8:58
• An infinite sum in the case of $\mathbb R$ would be an example of a countable input to $\square$, right? – user56834 Feb 10 at 8:59
• Also I can imagine that integration can be seen possibly in this context (in some interpretation it is a sum of an uncountable amount of infinitessimals) – user56834 Feb 10 at 9:02
• @user56834 With $(X,\cdot 1)=(\Bbb Z,+,0)$, what would the infinite "sum" $1+2+3+4+5+\ldots$ be? If you say it's $\infty$ (or even if you say it's $-\frac1{12}$) it is certainly not $\in X$. There is simply no agreaable-upon extension of the concept of sum to infinite sets unless in the cases of very special circumstances (e.g., all but finitely many summands are $=0$). There's a reason why infinite "sums" are actually called series – Hagen von Eitzen Feb 10 at 9:02
• Even in the reals most of them would not be defined. – Tobias Kildetoft Feb 10 at 9:03