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.

  • 2
    $\begingroup$ 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

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.