1
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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. $f_{S\cap T}(x)=f_S(x)f_T(x)$.

$\endgroup$
  • $\begingroup$ 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 May 5 '14 at 9:30
  • $\begingroup$ Yes. For the same reason some write $\cdot$ and $+$ instead of $\land$ and $\lor$ $\endgroup$ – Hagen von Eitzen May 5 '14 at 9:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.