Infimums of subsets of monoid. I came up with the following idea This may have already been studied by someone else or may be a common sense idea in this field.
Let $(M,+,\leq)$ be a totally ordered commutative monoid. For any non-empty subsets $A,B$ of $M$, we define a relation $A\precapprox B$ to be that for any $b\in B$, there is an element $a\in A$ such that $a\leq b$, similarly $A\approx B$ to be that $A\precapprox B$ and $B\precapprox A$. Then the relation $\approx$ is an equivalence relation. We denote the quotient $(\mathcal{P}(M)-\{\varnothing\})/\approx$ by $C_{\inf}(M)$ and the equivalence class of $A$ by $\inf A$.
I think that $C_{\inf}(M)$ has totally ordered commutative monoid structure as follows:

*

*$(\inf A)+(\inf B)=\inf (A+B)$, where $A+B=\{a+b\in M\mathrel{\vert} a\in A,b\in B\}$;


*$\inf A\leq \inf B$ if and only if $A\precapprox B$.
Presumably these operations are well-defined. Moreover
$$
M\longrightarrow C_{\inf}(M);x\longmapsto \inf \{x\}
$$
is a monomorphism of monoids. For example we have $C_{\inf}(\mathbb{N})=\mathbb{N}$ and $C_{\inf}(\mathbb{R})=\mathbb{R}\cup \{-\infty\}$. Has this kind of idea already been mentioned? If it has been studied, please provide references.
 A: Your idea makes sense and has actually been used in a more general setting. An ordered semigroup $(S, \leqslant)$ is a semigroup equipped with a partial stable order (that is, $u \leqslant v$ implies $sut \leqslant svt$).
Let $(S,\leqslant)$ be an ordered semigroup. A subset $U$ of $S$ is an upper set of $S$ if, whenever $x \in U$ and $x \leqslant y$, then $y \in U$. Given a subset $E$ of $T$, the upper set generated by $E$ is the set
$$
\uparrow\! E = \{t \in S \mid \text{there exists $s \in E$ such that $s \leqslant t$}\}
$$
Define the product of two upper sets $X$ and $Y$ as the upper set generated by $XY$, that is,
$$
 XY = \{ z \in S \mid \text{there exist $x \in X$ and $y \in Y$ such
 that $xy \leqslant z$} \}
$$
This operation makes the set $\cal{P}^\uparrow(S)$ of all upper sets of $S$ a semigroup. To get an ordered semigroup, define a relation $\leqslant$ on $\cal{P}^\uparrow(S)$ by setting $X \leqslant Y$ if and only if $Y \subseteq X$. Your approach works as well, in view of the following result.
Proposition. One has $X \leqslant Y$ if and only if, for each $y \in Y$, there exists $x \in X$ such that $x \leqslant y$.
Proof. Suppose that $Y \subseteq X$. Then the condition of the
statement is clearly satisfied by taking $x = y$. Conversely, suppose
that this condition is satisfied and let $y \in Y$. Then there exists
an element $x \in X$ such that $x \leqslant y$. Since $X$ is an upper set,
$y$ is also in $X$ and thus $Y$ is a subset of $X$.
Also note that, with your notation, $X \approx Y$ if and only if $\uparrow\! X = \uparrow\! Y$.
Dually, one could define the semigroup of lower sets of $S$. Please find below a few (advanced) papers where these notions are used.
[1] J. Almeida, A. Cano, O. Klíma and J.-É. Pin, Fixed points of the lower set operator, Internat. J. Algebra Comput. 25 (2015), 259-292.
[2] A. Cano and J.-É. Pin, Upper set monoids and length preserving morphisms, Journal of Pure and Applied Algebra 216, (2012), 1178-1183
[3] L. Polák, Operators on classes of regular languages, Semigroups, Algorithms, Automata and Languages (G. S. M. Gomes, J.-É. Pin, and P. Silva, eds.), World Scientific, 2002, pp. 407–422.
