Constructing a commutative monoid having idempotent elements (the underlying monoid of a Boolean ring) free over a set $X$, I arrive on a very natural way at monoid $M$ having the finite subsets of $X$ as underlying set and equipped with multiplication $ST:=S\cup T$. Any function $f:X\rightarrow\left|N\right|$ where $\left|N\right|$ denotes the underlying set of a monoid $N$ (again commutative and having idempotent elements) induces monoidmorphism $f^{\flat}:M\rightarrow N$ defined by $S\mapsto\prod_{s\in S}f\left(s\right)$.

What bothers me is that uptil now I did not encounter this in literature. Not $\cup$ is used as multiplication, but $\cap$. For instance on nlab I read under Boolean rings:

... as the free $\mathbb{Z}_{2}$-vector space $\mathbb{Z}_{2}$$\left[M_{n}\right]$ generated from the commutative idempotent monoid $M_{n}$ on $n$ generators. The latter can be identified with the power set on an $n$-element set with multiplication given by intersection, and $\mathbb{Z}_{2}$$\left[M_{n}\right]$ therefore has $2^{2^{n}}$ elements.

I understand that, especially if $X$ is a finite set, the monoids $\left(\wp\left(X\right),\cup\right)$ and $\left(\wp\left(X\right),\cap\right)$ are isomorphic ($S\mapsto S^{c}$) but a monoid-morphism $f^{\flat}:M\rightarrow N$ defined by $S\mapsto\prod_{s\in S^{c}}f\left(s\right)$ is less natural in my eyes. If it comes to infinite sets $X$ then cofinite subsets (I prefer finite subsets) come in sight.

Is there some underlying reason for the choice for intersection here?


1 Answer 1


As you note, the distinction disappears after switching to complements (and hence from finite to cofinite subsets). If one considers characteristic maps $f_S\colon X\to\{0,1\}$ with $$f(x)=\begin{cases}1&\text{if }x\in S\\0&\text{if }x\notin S\end{cases}$$ Then $\cap$ corresponds to pointwise multiplication, i.e. $f_{S\cap T}(x)=f_S(x)f_T(x)$.

  • $\begingroup$ Hmm... So the reason for preferring intersection is that $f_{S\cap T}\left(x\right)=f_{S}\left(x\right)f_{T}\left(x\right)$ is more 'convenient' than $f_{S\cup T}\left(x\right)=1-\left(1-f_{S}\left(x\right)\right)\left(1-f_{T}\left(x\right)\right)$? $\endgroup$
    – drhab
    Commented May 5, 2014 at 9:30
  • $\begingroup$ Yes. For the same reason some write $\cdot$ and $+$ instead of $\land$ and $\lor$ $\endgroup$ Commented May 5, 2014 at 9:56

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .