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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!

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    $\begingroup$ The wikipedia uses (3) but not (1) $\endgroup$
    – Keinstein
    Jul 30, 2019 at 14:55

2 Answers 2

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Actually you should differentiate between a dense order relation and a dense subset of an ordered set.

The order relation is dense when (3) holds.

A $Y$ is dense in an ordered set $(X,≾)$ iff a modified condition (1) holds: \begin{equation} ∀x,y∈X:x≺y,∃z∈Y:x≾z≾y \end{equation} This reads: There are two sets: One set $X$ which provides order relation and works as the universe, here. And another set $Y$ which has a nonempty intersection with every closed order interval. This is more or less a relational description of the same idea which is behind the notion of a dense set in topology: Every element of $X$ has sufficiently many elements of $Y$ nearby. Or in other words $Y$ may not be the full set $X$, but $Y$ is a sufficiently close approximation of $X$ in some setting. And yes, $X$ should be dense in $X$ in this sense.

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  • $\begingroup$ A slight modification for partial order $\precsim$: $\forall x,y\in X: x\prec y, \exists z \in Y: x\precsim z \precsim y$. $x$, $y$ should be distinct elements in $X$ to avoid the case $x=y\in X/Y$. When $x=y\in X/Y$, $x\succsim y$ is true (by reflexivity) but $\nexists z\in Y$, s.t. $x\precsim z \precsim y$ (by transitivity and anti-symmetry). $\endgroup$
    – Han Wang
    Feb 18, 2021 at 4:26
  • $\begingroup$ @HanWang you are right. I fixed the answer. $\endgroup$
    – Keinstein
    Feb 18, 2021 at 14:09
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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.

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  • $\begingroup$ If I interpret you correctly, then the definition should include either $z \notin \{x,y\}$ or $x \prec z \prec y$? But which (given that the second condition is much stronger than the first if we allow for $x \precsim z \precsim y$)? $\endgroup$ Sep 11, 2016 at 16:19
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    $\begingroup$ Allowing weak equality and requiring that $z\notin\{x,y\}$ is the same as requiring $x\prec z\prec y$. $\endgroup$
    – Asaf Karagila
    Sep 11, 2016 at 16:27
  • $\begingroup$ I was thinking that we could have $\forall x,y \in X$, $x \precsim y$ and $y \precsim x$ and allowing for $z \in X\backslash \{x,y\}$ satisfying the first and not the second, so that the two conditions would not, in general, be the same. Am I thinking wrong? Edit: I am. I'm thinking about preorders and not orders. I apologize. $\endgroup$ Sep 11, 2016 at 16:29
  • $\begingroup$ Thank you for your answer! $\endgroup$ Sep 11, 2016 at 16:32
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    $\begingroup$ The only reasonable way to define a dense preorder is to require that the induced partial order (taking a quotient by the obvious equivalence relation). $\endgroup$
    – Asaf Karagila
    Sep 11, 2016 at 16:32

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