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Let $X$ be an infinite set and $M$ a commutative monoid. Find a function $f \colon \mathcal{P}(X) \to M$ such that

  • $f(\emptyset) = 0$
  • for each element $x$ of $X$, $f(\{x\}) = 0$,
  • for any two disjoint subsets $A$ and $B$ of $X$, $f(A \cup B) = f(A) + f(B)$.

This has an obvious solution, the constant function $\mathcal{P}(X) \to \{0\}$. Is that the only solution? In particular, is that the only solution when $M$ is $\mathbf{N}$, $\mathbf{Z}$, $\mathbf{Q}$, or $\mathbf{R}$?

[Edited in light of Matthew Daly's answer]

A. Burrell
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    I'm not perfectly familiar with the context in which you are working, but it reminds me of probability measures and that $\Pr(X=x)=0$ for all $x$ for $X$ a continuous random variable despite $\Pr(X\in\Bbb R)=1$. The crux being that although we require $f(A\cup B)=f(A)+f(B)$ for finite and maybe even countable unions, this does not say anything about uncountable unions. – JMoravitz May 14 '21 at 15:04
  • See also https://math.stackexchange.com/q/28940 https://math.stackexchange.com/q/4024478 – Tom Collinge May 14 '21 at 15:10
  • @JMoravitz Defining the measure on the whole powerset doesn't seem trivial, though (and is impossible for the Lebesgue measure for example). – Klaus May 14 '21 at 15:12
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    @JMoravitz The context is defining summation operators, the things generally written $\sum$. For a commutative monoid $M$, and some set $I$ of possible indices, the summation operator can be thought of as a mapping from the set of finitely additively supported $M$-valued families, indexed by subsets of $I$, to $M$. The operator can be defined as the unique mapping satisfying certain conditions; how strong do those conditions need to be? In particular, is it necessary to require that all families with empty support map to the identity element? – A. Burrell May 14 '21 at 16:21

2 Answers2

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Let $M=\{0,1\}$ with $0+0=0$ and $1+0=0+1=1+1=1$. This is a commutative monoid.

Then define $$f(S) = \begin{cases} 0 & S\text{ is finite} \\ 1 & S \text{ is infinite}\end{cases}$$

This satisfies the conditions because the empty set, singletons, and the union of two finite sets are all finite.

  • Thanks for the response. Besides answering, very neatly, the question in the title, it suggests that the answer to the question in the body might be, it depends on the monoid. I've edited the question in light of this. – A. Burrell May 14 '21 at 16:40
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If your monoid has only one element $0$, then obviously that's the only solution. Otherwise "nonprincipal ultrafilters" give you a solution when the monoid has at least two elements $0$ and $a$.

A filter on a set $X$ is a collection $\mathcal{F} \subseteq \mathcal{P}(X)$ such that:

  • $X \in \mathcal{F}$, $\emptyset \notin \mathcal{F}$
  • $\forall A, B \subseteq X (A \subseteq B \wedge A \in \mathcal{F} \rightarrow B \in \mathcal{F})$
  • $\forall A,B \in \mathcal{F} (A \cap B \in \mathcal{F})$

A filter is an ultrafilter if for every $A \subseteq X$ either $A \in \mathcal{F}$ or $X \setminus A \in \mathcal{F}$. It is non-principal if it contains no singletons.

Non-principal ultrafilters exist on any infinite set $X$. So if $\mathcal{F}$ is such a filter on $X$, then mapping $f(A) = 0$ if $A \notin \mathcal{F}$, $f(A) = a$ if $A \in \mathcal{F}$, gives you a non zero solution.

Jonathan Schilhan
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