Commutative monoid with natural order is a poset

Question

Let $(M,+,0)$ be a commutative monoid and write for $x,y\in M$, $x\leq y \iff \exists t\in M: x+t=y$.

I want to show $(M,\leq)$ is a poset.

I am stuck at showing antisymmetry.

Obviously $x+0=x \Rightarrow x\leq x$. And if $x\leq y$ and $y\leq z$ then $\exists a,b\in M$ with $$x+a=y \wedge y+b=z \Rightarrow x+(a+b)=y+b=z \Rightarrow x\leq z.$$ For antisymmetry: $x+a=y \wedge y+b=x$ then $y+b=x+a+b=x$ so $$x=x+a+b$$ But I cannot get any further.

Answer 1

This is not true in general. For a reference see Lemma 5.1.4 in the notes on Blueprints by Oliver Lorscheid and the example following it. You can find it here http://lorschei.impa.br/2018-Blueprints/lecturenotes.php

Answer 2

As yamete kudasai states, the reason you can't prove this is because it isn't true. You don't need to go read some notes to figure this out, though. Just consider an example. In fact, almost any (commutative) group will do. For a group, we have $g\leq h$ for all $g$ and $h$ via $t=h-g$. Antisymmetry would then imply $g=h$ for all elements of the group.

I assume you've established reflexivity and transitivity, and a set equipped with relation satisfying those is called a preordered set, sometimes abbreviated to a proset.