0
$\begingroup$

Given:

  • A set $M$.
  • A binary operation $+$ defined on $M$
    $+: M \times M \to M$
    $\text{ that is both associative and commutative.}$

satisfying the following properties:

P-1: $\text{For every } x,y,z \in M \text{, if } z + x = z + y \, \text{ then } \, x = y$.

P-2: $\text{For every } x,y,z \in M \text{, if } z = x + y \, \text{ then } \, z \ne x$.

P-3: $\text{For every } x,y \in M \text{, if } x \ne y \, \text{ then } \, [\exists u \; | \, x = y +u] \text{ or } [\exists u \; | \, y = x +u]$.

Example: The set of positive real numbers.

Are there examples where the the cardinality of $M$ is strictly greater that $|\mathbb R |$?

$\endgroup$
  • 2
    $\begingroup$ Keywords: first-order theory, Löwenheim–Skolem, ultrapower $\endgroup$ – bof Oct 19 '18 at 7:48
  • 1
    $\begingroup$ My first thought was this: Taking your example $(\mathbb{R}^+,+)$, let $M$ denote the set of all functions $f : \mathbb{R} \to \mathbb{R}^+$. An addition $+$ on $M$ is as usual defined by $(f + g)(x) = f(x) + g(x)$. Unfortunately axiom 3 is not satisfied. $\endgroup$ – Paul Frost Oct 19 '18 at 12:29
2
$\begingroup$

You can get such a semigroup by taking the set of positive elements in any totally ordered abelian group. Now, if $A$ is a totally ordered abelian group and $S$ is any totally ordered set, the direct sum $A^{\oplus S}$ is also a totally ordered group with respect to the lexicographic order. Explicitly, the semigroup of positive elements of $A^{\oplus S}$ is the set of functions $f:S\to A$ such that $f(s)=0$ for all but finitely many $s\in S$ and $f(s)>0$ for the least $s\in S$ for which $f(s)$ is nonzero. If $A$ is nontrivial then this semigroup has at least as many elements as $S$, so you can get an example of arbitrarily large cardinality by taking $S$ to be a totally ordered set of arbitrarily large cardinality.

Much more generally, any theory over a countable first-order language which has an infinite model has models of all infinite cardinalities, by the Löwenheim-Skolem theorem. Your semigroups are just models of a certain first-order theory over the language with a single binary operation $+$.

$\endgroup$

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.