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Give three finite multisets $A,B,C$ I am trying to devise a sufficient condition that implies equality of both $A$ and $B$ with $C$. More formally, if $A,B,C$ are defined over a space $\Omega$, I am looking for a function $f : \Omega \times \Omega \to \Omega$ such that $$f(A,B) = C, \text{ implies that } A = C \text{ and } B = C.$$ Naturally, equality is as multisets.

The motivation for this question was noticing the basic observation that if $A\cup B \subseteq C$ as multisets, then $A \subseteq C$ and $B \subseteq C$ as multisets. I was wondering if something similar happens with the equality "$=$" operator instead of the subset "$\subseteq$" operator. In the subset example, the function $f$ is the union of multisets $\cup$ defined as $$f(A,B) = A \cup B := \{[x,\max\{m_A(x),m_B(x)\}] : x \in A \text{ or } x \in B\},$$ where $m_Z(x)$ counts the number of times that $x$ appears in some multiset $Z$.

Bean Guy
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  • This is very hard to follow. What does "the equality of the first two with the third" mean? And what "subsets" are you talking about"? – lulu Feb 17 '23 at 14:38
  • Can you solve your problem for ordinary Sets? – lulu Feb 17 '23 at 14:40
  • @lulu Fixed the typos – Bean Guy Feb 17 '23 at 14:41
  • I still don't understand. Is $f$ only defined on the diagonal? Otherwise, if $A\neq B$ then what is $f(A,B)$ supposed to mean? – lulu Feb 17 '23 at 14:46
  • To stress: this issue doesn't apply for the ordinary inclusion. Given any two sets, or multisets, there is a "least" set or multiset that contains both. But of course it is not true that given any two multisets, that there is a multiset equal to both. – lulu Feb 17 '23 at 14:48
  • @lulu $f$ is just a function that takes two multisets and produces a third one. So, for example $f$ could be intersection or union of multisets (or a finite combination of some of them). – Bean Guy Feb 17 '23 at 14:49
  • Well, in that case, given any two multisets $A,B$ you could always conclude that $A=B=f(A,B)$. Since this is evidently absurd, no such function exists. – lulu Feb 17 '23 at 14:51

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To summarize the discussion in the comments:

No such function can exist (at least, assuming $\Omega$ is non-empty).

To see this, suppose we had such a function and let $\omega \in \Omega$ be some element, with $A=\{\omega\}$. Then we consider $f(A,\emptyset)$ as some set (or multiset). By definition we must have $A=\emptyset=f(A,\emptyset)$ so, in particular, $A=\emptyset$, counter to the definition of $A$, and we have a contradiction.

Technical point: Writing $f:\Omega\times \Omega\to \Omega$ can't be what was intended. That would mean a function that took two elements from $\Omega$ and returned a third. I assume that what was meant was a function, or more likely a functor, from $\mathbf {MSets}_{\Omega}\times \mathbf {MSets}_{\Omega}\to \mathbf {MSets}_{\Omega}$. That is, something that takes two multisets (of elements taken from $\Omega$) and returns a third. That's what the contradiction I described assumes.

lulu
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