Can a certain monoid exist?

Question

Is it possible to have an uncountable commutative monoid, where for every $a$ in the monoid, $a+a=a$?

I have a set which I am trying to define a group structure on (I am settling for a monoid structure though). the set is a bunch of equivalence classes of the real numbers, and I'm trying to see if it would be possible to define addition of classes by addition of representatives. The problem is adding an element of any class to itself, will always get you in the same class, so i'm not sure if this would mean I cant define the structure this way.

Also I should add, when I say i'm defining addition by addition of representatives, each class has to have a canonical representative which is used in the addition operation, as addition definitely is dependant on the choice of representative.

Answer 1

Here is some background. An element satisfying $a + a = a$ is called idempotent. A commutative monoid in which every element is idempotent is called, naturally, an idempotent commutative monoid.

These things are plentiful, and in fact are more or less equivalent to (bounded) (join-)semilattices, which are partially ordered sets $(P, \le)$ with a bottom / least element $\bot$ (an element such that $\bot \le p$ for all $p \in P$) and binary joins $p \vee q$ (an element such that $p \le x$ and $q \le x$ iff $p \vee q \le x$). Joins are always commutative and idempotent, $\bot$ is the identity element, and the partial order can be recovered from the join operation since $p \le q$ iff $p \vee q = q$ (exercise).

Joins are a generalization of the $\text{max}$ operation; here are a few illustrative examples.

• The nonnegative real numbers $\mathbb{R}_{\ge 0}$ are totally ordered (so in particular partially ordered). The bottom element is $0$ and the join is $\text{max}$.
• If $X$ is any set, the set $P(X)$ of subsets of $X$ is partially ordered by inclusion. The bottom element is the empty subset, and the join is the union.
• The set $\mathbb{N}$ of positive integers is partially ordered by divisibility. The bottom element is $1$, and the join is $\text{lcm}$.
• If $V$ is any vector space, the set of subspaces of $V$ is partially ordered by inclusion. The bottom element is the zero subspace, and the join is the sum.

The first, second, and fourth examples all give examples of uncountable idempotent commutative monoids; the second and fourth examples in particular admit many variations. For example one can consider subgroups of a group, subrings of a ring, submodules of a module, etc.

Answer 2

Sure. Take $M = [0..1)$ within $ℝ$ and as operation taking maxima $$∨~ \colon M × M → M,~(x, y) ↦ \max (x,y).$$ This yields an uncountable, commutative monoid with zero as identity and $a ∨ a = a$ for all $a ∈ M$.