I am trying to understand a proposition which goes like this:

Let $(X,\precsim)$ be a nonempty totally preordered set.

Let $(\hat{X}, \precsim)$ be the set of all equivalence classes in $(X,\precsim)$, i.e. $X$'s quotient set endowed with the same order as $X$ (thus clearly a totally-ordered set).

Then $\exists f: \hat{X} \rightarrow \mathbb{R}$, where $f$ is order-embedding, iff $\exists X^* \subset \hat{X}$ such that $X^*$ is at most countable and order-dense (close-packed) in $\hat{X}$. ($\mathbb{R}$ is ordered as usual).

I just can't understand how to prove this nor get any intuition why this is so. (The fact that it starts with a proset and not a toset is irrelevant to my lack of understanding, I just wanted to state it in full.)

Edit: Order-embedding, not order-isomorphic. Sorry for the confusion. (bis)

  • $\begingroup$ I'm not quite sure I understand your claim, since there seems to be no reason that the existence of such $X^\ast$ should even imply that $\hat{X}$ and $\mathbb{R}$ even have equal cardinality, so certainly cannot get an order-isomorphism (=bijection which is order-preserving and order-reflecting). For example, take $\hat{X}=X^\ast = \mathbb{Q}$. $\endgroup$
    – Hayden
    Sep 11 '16 at 5:47
  • $\begingroup$ It is not my claim. Basically, it claims that for some set $X$ and some total order $\precsim$, $(X,\precsim)$ is isomorphic wrt $\mathbb{R}$ iff there exists some subset of $X$ which is countable and order-dense in $(X,\precsim)$. I know I'm repeating myself, but I can't put it clearer than this. My question is how to prove this or, at least, why this should be so (I'm admitting that it is so) $\endgroup$ Sep 11 '16 at 5:51
  • $\begingroup$ Sure, and I'm saying it doesn't appear to be true, with the counter-example of $(\mathbb{Q},\leq)$, unless I'm misunderstanding something in the claim (which I will continue to call a claim since it is unproven). $\endgroup$
    – Hayden
    Sep 11 '16 at 5:54
  • $\begingroup$ I apologize. Order-REFLECTING not isomorphic. My problem is that I'm trying to translate an economics theorem to its proper math language $\endgroup$ Sep 11 '16 at 5:56
  • $\begingroup$ thanks for the clarification, this seems much more likely to be true (perhaps even an order-embedding). $\endgroup$
    – Hayden
    Sep 11 '16 at 5:56

First let me restate the proposition (as I understand it) in human language, rather than try to imitate your notation.

Proposition. Let $X$ be a totally ordered set. The following statements are equivalent:
(1) there is a set $D\subseteq X$ which is (at most) countable, and which is dense in $X,$ in the sense that $D\cap[a,b]\ne\emptyset$ whenever $a,b\in X,\ a\lt b.$
(2) $X$ is order-isomorphic to a subset of $\mathbb R.$

$\underline{\text{(1)}\implies\text{(2)}}:$ Let $D=\{d_n:n\in\mathbb N\}.$ Define $f:X\to\mathbb R$ by setting $$f(x)=\sum_{d_n\lt x}2^{-n}+\sum_{d_n\le x}2^{-n}.$$ It is easy to see that $x,y\in X,\ x\lt y\implies f(x)\lt f(y).$

$\underline{\text{(2)}\implies\text{(1)}}:$ (Axiom of choice needed here.) Without loss of generality, we assume that $X\subseteq\mathbb R.$ For each rational number $r,$ let $x_r$ be the greatest element of $X\cap(-\infty,r)$ if such exists, and let $y_r$ be the least element of $X\cap(r,\infty)$ if such exists. For each rational interval $(r,s)$ such that $X\cap(r,s)\ne\emptyset,$ choose an element $z_{r,s}\in X\cap(r,s).$ Let $D$ be the set of elements $x_r,y_r,z_{r,s}$ so chosen. Clearly $D$ is a countable dense subset of $X.$

  • $\begingroup$ Great! Thank you! Now I just need the $\Leftarrow$! (Can't you relax the inequality here?) $\endgroup$ Sep 11 '16 at 6:19
  • $\begingroup$ Did I interpret the question correctly? What inequality do you want relaxed? $\endgroup$
    – bof
    Sep 11 '16 at 6:26
  • $\begingroup$ You did interpret it correctly. The inequality I was mentioning is right above $(2) \Longrightarrow (1)$, i.e. "It is easy to see..." $\endgroup$ Sep 11 '16 at 6:47
  • 2
    $\begingroup$ For (2) $\implies$ (1), we assume that $X$ is order-isomorphic to a subset $X'$ of $\mathbb R.$ In order to prove that $X$ has a countable dense subset, it suffices to show that $X'$ has a countable dense subset, since $X$ is isomorphic to $X'.$ That's why there is no loss of generality is assuming that $X$ is a subset of $\mathbb R.$ $\endgroup$
    – bof
    Sep 11 '16 at 7:02
  • 1
    $\begingroup$ I understood now your answer. Thank you for your patience and for your answer! $\endgroup$ Sep 11 '16 at 7:20

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