The usual definitions of order-dense set I've seen are:
(1) Let $\precsim$ be an order defined on set $X$. $(X,\precsim)$ is order-dense if $\forall x,y \in X: x \precsim y, \exists z \in X: x \precsim z \precsim y$ (wikipedia, math.se).
(2) Let $\precsim$ be an order defined on set $X$. $(X,\precsim)$ is order-dense if $\forall x,y \in X: x \prec y, \exists z \in X: x \precsim z \prec y$ (some lecture notes).
(3) Let $\precsim$ be an order defined on set $X$. $(X,\precsim)$ is order-dense if $\forall x,y \in X: x \prec y, \exists z \in X: x \prec z \prec y$ (proofwiki, math.se).
where $\precsim$ is a weak partial or total order and $\prec$ denotes the strict version of it.
Although similar, these are quite different. In sum, some definitions require some or all relations between the elements to be strict ($\prec$ instead of $\precsim$) and others even specify that $z \in X \backslash \{x,y\}$.
Is there a consensual definition of an order-dense set?
- Can $z \in X$ or has to be distinct from $x,y$, i.e. $z \in X \backslash \{x,y\}$?
- Can $x\precsim z \precsim y$ or has to be $x\prec z \prec y$?
(This is not trivial, as (1) holds trivially for any partially-ordered set; (2) and (3) prevent $X$ to be a singleton and, requiring $z \in X \backslash \{x,y\}$, would prevent $(\mathbb{N},\leq)$ to be order-dense in itself)
Thank you in advance!
In both (1) and (2) taking $x=z$ will satisfy the conclusion, making every order a dense order. Unless we require that $z\notin\{x,y\}$, or a condition which implies that like $x\prec z\prec y$, there is no way to guarantee that some orders will not be dense.
While it is somewhat unusual for a singleton to be a "dense set", the definition does make sense if you only require that when $x\prec y$, then there is some $z$ such that $x\prec z\prec y$. Therefore often you should also require that for every $x$ there is some $y$ such that $x\prec y$ or $y\prec x$. Otherwise taking a discrete ordering will be dense vacuously.