On the powerset of a monoid, does the second operation play nice with the first? Indeed, is it even useful?

Question

Let $X$ denote a monoid. Then we can make $Y = \mathcal{P}(X)$ into a monoid, too. Define $$AB = \{ab \mid a \in A, b \in B\}$$ for all $A,B \in Y.$ We see immediately that $1$ (shorthand for $\{1\}$) is our new identity:

$$A1 = A, \;\; 1B = B.$$

In fact, $Y$ becomes an ordered monoid by defining that $A \leq B$ is notation for $A \subseteq B$. It follows that:

• $A \leq B \rightarrow AC \leq BC$
• $A \leq B \rightarrow CA \leq CB$

Furthermore, $Y$ is a complete atomistic Boolean algebra, and we have compatibility of composition with joins:

• $A \left(\bigvee_{i \in I} B_i\right) = \bigvee_{i \in I} AB_i$

• $\left(\bigvee_{i \in I} A_i\right) B = \bigvee_{i \in I} A_i B$

Now most authors would probably stop there. And perhaps that is the right thing to do. But, for the sake of experimentation, lets go a step further. Define another monoid structure on $Y$ by writing

$$A*B = (A^cB^c)^c.$$

Question. Does this new operation "play nice with" the earlier-defined operation in some sense? Indeed, is the $*$ operation in any way useful?

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Discussion. We see that $*$ is associative, and that $1^c$ is its identity

$$A * 1^c = A, \;\; 1^c * B = B.$$

We also have the following.

• $A \leq B \rightarrow A*C \leq B*C$

• $A \leq B \rightarrow C*A \leq C*B$

• $A * \left(\bigwedge_{i \in I} B_i\right) = \bigwedge_{i \in I} (A*B_i)$

• $\left(\bigwedge_{i \in I} A_i\right) * B = \bigwedge_{i \in I} (A_i * B).$

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Remark. We can do something similar with the binary relations on a set. Given binary relations $\alpha$ and $\beta$ on a set $S$, define

$$\alpha\beta = \{(x,y) \in S^2 \mid \exists s \in S : (x,s) \in \alpha \wedge (s,y) \in \beta\}$$

$$\alpha * \beta = \{(x,y) \in S^2 \mid \forall s \in S : (x,s) \in \alpha \vee (s,y) \in beta\}.$$

This makes $\mathcal{P}(S^2)$ into an ordered monoid in two distinct ways. In the first way, the diagonal relation is the identity. In the second, its complement is the identity. All the expected interactions with order-theoretic concepts hold. Actually, we can go further. In particular, the class of all sets can be made into a category whose morphisms are binary relations. However, this can be done in two different ways, corresponding to two different laws of composition. Once again, all the expected interactions with order-theoretic concepts hold.

Answer 1

Too many questions in the one. I answer on the first:

The new monoid $\mathcal{P}(X)(*)$ is isomorphic to $\mathcal{P}(X)$ by the map $A\to A^c$ so probably it is not useful.

Answer 2

This does not really answer your question, but it is something you might find interesting: the powerset of a monoid and the set of binary relations on a set are examples of residuated lattices. These examples are explained in detail in the book Residuated Lattices: An Algebraic Glimpse at Substructural Logics by Galatos, Jipsen, Kowalski and Ono.

In the case of the set $A=\mathcal P(S\times S)$ of binary relations on a set $S$, we have two unary operations on $A$:

• the converse of $\alpha\in A$ is the relation $\alpha^{\mathrm{op}}$ defined by $(x,y)\in\alpha^{\mathrm{op}}\iff(y,x)\in\alpha$; and
• the complement of $\alpha\in A$ is the relation $\alpha^{\mathrm c}$ defined by $(x,y)\in\alpha^{\mathrm c}\iff(x,y)\notin\alpha$.

We can then define a binary operation $\backslash$ on $A$ by setting $\alpha\mathop\backslash\beta=(\alpha^{\mathrm{op}}\beta^{\mathrm c})^\mathrm c$. This operation satisfies the 'residuation' identity $$ \alpha\gamma\leq\beta\iff\gamma\leq\alpha\mathop\backslash\beta $$ for all $\alpha,\beta,\gamma\in A$, or in other words, for each $\alpha\in A$ the functions $\alpha\mathord-\colon A\to A$ and $\alpha\mathop\backslash\mathord-\colon A\to A$ constitute a (monotone) Galois connection.