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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.

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    $\begingroup$ This is utter nonsense. You cannot define $\square$ on infinite sets at all, countable or uncountable. $\endgroup$ – Eric Wofsey Feb 10 at 8:58
  • $\begingroup$ An infinite sum in the case of $\mathbb R$ would be an example of a countable input to $\square$, right? $\endgroup$ – user56834 Feb 10 at 8:59
  • $\begingroup$ 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) $\endgroup$ – user56834 Feb 10 at 9:02
  • $\begingroup$ @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 $\endgroup$ – Hagen von Eitzen Feb 10 at 9:02
  • $\begingroup$ Even in the reals most of them would not be defined. $\endgroup$ – Tobias Kildetoft Feb 10 at 9:03

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