Must an infinite sum of zeros be zero?

Question

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]

Answer 1

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.

Answer 2

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.