Basic concepts question.


I am used to the Cartesian product of two sets: $A \times B = \{(a,b) \mid a \in A, b \in B\}$.


Is there an operator that produces sets instead of tuples? We might call it *set-product* and define $A \otimes B := \{a \cup b \mid a\in A, b\in B\}$.


What I am actually trying to do is "factorize" a given set-family $\mathcal M \subseteq 2^V$ where $V$ is the base set. That is, I would like to write $\mathcal M = M\_1 \otimes M\_2 \otimes M\_3$ with $M\_i \subseteq 2^V$.


*example*:


* $\mathcal M = \{\{1,2\}, \{1,3\}, \{1,6\}\}$ can be factorized into $\{\{1\}\} \otimes \{\{2\},\{3\},\{6\}\}$


**My questions are:**


* do people use this operator (if so, what is it called? I don't want to reinvent the wheel)?
* do you know of results relating to the factorizability of sets in this way?


My goal is to connect what I am doing to existing work.


**EDIT**:


* fixed errors in question and definition of $\otimes$



Cartesian products and set-products
As far as I understand, you start with a "base set" $V$ and consider the monoid obtained by endowing the power set of the power set of $V$ with the binary operation that sends a pair $(A, B)$ of families of subsets of $V$ to $\{X \cup Y \colon X \in A,\, Y \in B\}$, which is still a family of subsets of $V$.
If my understanding is correct (is it?), then you may want to read about *power monoids*, especially if you are interested in questions relating to factorization (as it seems that you are).


Given a (multiplicatively written) monoid $H$, the *full power monoid* of $H$ is the monoid $\mathcal P(H)$ obtained by equipping the collection of all non-empty subsets of $H$ with the operation of setwise multiplication induced by $H$, so that $AB := \{ab: a \in A, \, b \in B\}$ for all $A, B \in \mathcal P(H)$. The OP is about the special case where $H$ is the monoid obtained by endowing the power set of a certain set $V$ with the binary operation $(X, Y) \mapsto X \cup Y$. So the answer to the question:



> 
> do you know of results relating to the factorizability of sets in this way?
> 
> 
> 


is yes in the sense that this is a special case of a broader class of monoids where people (including myself) have studied, and are studying, various properties that can be labelled with the "Factorization" tag. However, you may want to consider that, before even starting to tackle the basic question of the existence of a factorization for certain elements of the monoid you're implicitly considering (and, more generally, of any monoid), you should somehow decide which "building blocks" are allowed in the factorization process, which is a (delicate) question of independent interest.


From the perspective of the classical theory (of factorization), the "canonical choice" would be to use as building blocks either the atoms or the irreducibles of the monoid $M$ under consideration, where an *atom* is a non-unit $a \in M$ such that $a \ne xy$ for all non-units $x, y \in M$ (this definition is commonly attributed to P.M. Cohn), and an irreducible is a non-unit $a \in M$ such that $a \ne xy$ for all $x, y \in M$ such that $MxM \ne MaM \ne MyM$ (in the commutative case, this definition is commonly attributed to D.D. Anderson & S. Valdes-Leon). However, none of these choices is fit for your case, unless $V$ is a finite set.


If, on the other hand, you're actually interested in the case where the elements of your monoid are *non-empty, finite* families of subsets of $V$ (are you?), then things are quite different and, with the same notation as above, you would better move from the *full* power monoid $\mathcal P(H)$ to the *power monoid* $\mathcal P\_{\rm fin}(H)$ of $H$, that is, the submonoid of $\mathcal P(H)$ whose elements are the (non-empty) finite subsets of $H$: It is obvious that $\mathcal P\_{\rm fin}(H)$ is strongly $\subseteq$-artinian, meaning that the $\subseteq$-height of every $X \in \mathcal P\_{\rm fin}(H)$ is finite; as a result, every $X \in \mathcal P\_{\rm fin}(H)$ factors into a (finite) product of $\subseteq$-irreducibles (see the notes below). Then, you are left with the problem of characterizing the $\subseteq$-irreducibles. (You could also use other preorders, but the inclusion order is a natural go-to here.)


By the way, product-sets in all sorts of monoids (and, in particular, sumsets in additively written groups) are one of the main objects of study in [arithmetic combinatorics](https://en.wikipedia.org/wiki/Arithmetic_combinatorics). So, stitching the pieces together, I would say that also the answer to the question:



> 
> do people use this operator (if so, what is it called? I don't want to reinvent the wheel)?
> 
> 
> 


is yes (and you are reinventing the wheel).


**Notes.** Given a preorder $\preceq$ on a monoid $M$, a *$\preceq$-non-unit* of $M$ is an element $u \in M$ that is not $\preceq$-equivalent to the identity $1\_M$ (two elements $x, y \in M$ are *$\preceq$-equivalent* if $x \preceq y \preceq x$). Then, the *$\preceq$-height* of an element $x \in M$ is the supremum of the set of all integers $n \ge 1$ for which there exist $\preceq$-non-units $x\_1, \ldots, x\_n \in M$ with $x\_1 = x$ and $x\_{i+1} \prec x\_i$ for $1 \le i < n$, with the understanding that $\sup \emptyset := 0$; and a *$\preceq$-irreducible* of $M$ is a $\preceq$-non-unit $a \in M$ such that $a \ne xy$ for all $\preceq$-non-units $x, y \in M$ with $x \prec a$ and $y \prec a$.


