3
$\begingroup$

This a follow up question of Chain components and posets.

Let $X$ be a compact metric space and $R\subset X^2$ a closed, transitive relation. Denote by $|R|=\{x\in X: xR x\}$ the diagonal of $R$. The relation $x\sim y \Leftrightarrow x R y\ \land\ y R x $ is an equivalence relation on $|R|$ and the equivalence classes form a poset if we say that $A\le_R B$ for $A,B\in |R|/_{\sim}$, if $x Ry$ for some $x\in A$ and $y\in B$.

Is there an explicit characterization of the posets that can/cannot arise in this way given the metric space $X$?

$\endgroup$

1 Answer 1

1
$\begingroup$

It is easy to see that the posets $(U,\le)$ that can arise in this way are exactly such that for all $u$ and $v$ in $U$ $$u\le v\ \&\ v\le u\implies u=v. \tag{1}\label{1}$$ That is, any poset can arise this way.


Indeed, if for some $A$ and $B$ in $|R|/_{\sim}$ we have $A\le_R B\le_R A$, then for some $a,a_1$ in $A$ and some $b,b_1$ in $B$ we have $$aRb \quad \text{and}\quad bRb_1\!Ra_1\!Ra,$$ so that $aRb$ and, by the transitivity, $bRa$, so that $a\sim b$. So, $A\le_R B\le_R A$ implies $A=B$.

Vice versa, if \eqref{1} holds for a poset $(U,\le)$, let $R$ be $\le$. Then for any $u$ and $v$ in $U$ we have $u\sim v$ iff $u=v$ and hence $(|R|/_{\sim},\le_R)$ coincides with $(U,\le)$.


The conditions that $X$ is a compact metric space and $R$ is closed play no role here.

$\endgroup$
5
  • $\begingroup$ I'm sorry, I'm confused. Apart from other things, there has to be some connection with the properties of $X$. A trivial instance: if $X$ is finite, no infinite poset can arise, for instance. So how can it be that "any" poset can arise? $\endgroup$ Commented Jun 27 at 15:32
  • $\begingroup$ Also, I don't think that you can get a countable dense total order if $X$ is uncountable, if $R$ is closed. $\endgroup$ Commented Jun 27 at 15:33
  • 2
    $\begingroup$ @MarcoFarotti : Did you mean to assume that $X$ is fixed? From your "arise in this way" that would be hardly possible to guess. If you did mean that, you should have said "arise in this way given the metric space $X$". $\endgroup$ Commented Jun 27 at 15:43
  • $\begingroup$ Also, do not use $\mathcal R$ where $R$ will certainly do. $\endgroup$ Commented Jun 27 at 15:44
  • $\begingroup$ Yes, you're right. I edited the question, thank you. $\endgroup$ Commented Jun 27 at 16:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .