Examples for preordered algebraic structures, whose algebraic operations are not entirely compatible with the preorder

Question

Are there any interesting, useful or well-known examples for

1. an algebraic structure of the form $(X,\leq,+,0)$, where $X$ is a set containing $0$, $\leq$ is a preorder on $X$, $+$ is a binary operation on $X$, and $0$ is a neutral element w.r.t. $+$, such that $+$ is not entirely compatible with $\leq$ in the sense that the following condition does not hold for every $x,x',y,y'\in X$ (but may hold for some): $$ x\leq x'\ \wedge\ 0 \leq y\leq y'\ \implies x+y\leq x'+y'? $$

2. an algebraic structure of the form $(X,\leq,\cdot)$, where $X$ is a non-empty set, $\leq$ is a preorder on $X$, and $\cdot:(\mathbb{R}_+\times X)\rightarrow X$ is an operation that is not entirely compatible with $\leq$ in the sense that the following condition does not hold for every $r \in \mathbb{R}_+$ and every $x,x'\in X$ (but may hold for some): $$ x\leq x'\ \implies\ rx\leq rx'? $$

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* $\mathbb{R}_+$ is the set of non-negative real numbers.

Answer 1

Answer to the first question. Green's preorders on monoids are famous examples of preorders that are not in general compatible with the product. Let $M$ be a monoid.

\begin{align*} s \mathrel{\leqslant}_\cal{R} t &\text{ if and only if $s = tu$ for some $u \in M$}\\ s \mathrel{\leqslant}_\cal{L} t &\text{ if and only if $s = ut$ for some $u \in M$}\\ s \mathrel{\leqslant}_\cal{J} t &\text{ if and only if $s = utv$ for some $u, v \in M$}\\ s \mathrel{\leqslant}_\cal{H} t &\text{ if and only if $s \mathrel{\leqslant}_\cal{R} t$ and $s \mathrel{\leqslant}_\cal{L} t$} \end{align*} Equivalently, these preorders can be defined in terms of ideals\begin{align*} s \mathrel{\leqslant}_\cal{R} t & \text{ if and only if $sM \subseteq tM$}\\ s \mathrel{\leqslant}_\cal{L} t &\text{ if and only if $Ms \subseteq Mt$}\\ s \mathrel{\leqslant}_\cal{J} t &\text{ if and only if $MsM \subseteq MtM$}\\ \end{align*}