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Let $X$ be a countable simple ordered set. I am trying to prove that there exists on X a real, order-preserving function, continuous in the order topology.

I construct by induction an order-preserving function $f$ taking X into some finite real interval $[a,b]$. And I try to prove its continuity by prove $\{x\in X|f(x)\le \alpha, \alpha\in[a,b]\}$ is close. If $\alpha\in f(X)$ or $\{x\in X|f(x)<\alpha\}$ has a largest element, it's obvious. But it's an open set when $\{x\in X|f(x)<\alpha\}$ doesn't have a largest element. So I think that the function need to be modified. But I don't know how to do it.

Or such function may not exist? Is there a counterexample? Thanks for any help in advance.

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  • $\begingroup$ The constant map is order preserving and continuous. Do you mean order embeding? Yes every countable linear order, order embeds into the rationals. For continuity I'd consider the inverse map of (a,b). $\endgroup$ – William Elliot May 12 '17 at 19:03
  • $\begingroup$ Re: The A by Keith Kearnes: It is a theorem of Georg Cantor that any two countably infinite linear orders that are order-dense (i.e. if $a<b$ there exists $c$ with $a<c<b$), and which have no end-points, are order-isomorphic to each other. Hence,( since $\mathbb Q$ is a linear order of this kind), any such linear order is order-isomorphic to $\mathbb Q$. $\endgroup$ – DanielWainfleet May 13 '17 at 6:26
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[Original question]
Let X be a countable simple ordered set. I am trying to prove that there exists on X a real, order-preserving function, continuous in the order topology.

The question of whether a given countable linearly ordered set has a continuous embedding into $\mathbb R$ has an affirmative answer.

[Comment from above]
Yes every countable linear order, order embeds into the rationals.

It is true that every countable linear order embeds into the rationals, but such embeddings need not be continuous in the order topology.

Try this instead. I will use $L$ instead of $X$ to denote the countable linear order of the problem. Say that $b$ covers $a$ in $L$ if $a<b$ and the interval $[a,b]$ equals $\{a, b\}$; i.e., if there is no element between $a$ and $b$. Define $L^*$ to be the linear order obtained from $L$ by inserting a copy of $\mathbb Q$ (the rationals) between $a$ and $b$ whenever $b$ covers $a$, inserting a copy of $\mathbb Q$ at the bottom of $L$ if $L$ has a least element, and inserting a copy of $\mathbb Q$ at the top of $L$ if $L$ has a greatest element. (Roughly: fill in all gaps of $L$ with copies of $\mathbb Q$ to obtain a countable dense linear order without endpoints, $L^*$.) Let $\alpha\colon L\to L^*$ be inclusion.

$L^*$ is order isomorphic to the rationals, so there exists an order embedding $\beta\colon L^*\to \mathbb R$ which maps $L^*$ ONTO the rationals. I claim that the composite embedding $\gamma = \beta\circ\alpha \colon L\to L^*\to \mathbb R$ is continuous in the order topology.

To show continuity, it suffices to show that $\gamma^{-1}((r,\infty))$ and $\gamma^{-1}((-\infty,s))$ are open in $L$ for each real $r$ and $s$. By symmetry it is enough to just show that $\gamma^{-1}((r,\infty))$ is open. To obtain a contradiction, assume that $F=\gamma^{-1}((r,\infty))$ is not open. Since $F$ is an upward closed subset of $L$ that is not open, $F$ must have a least element (say $t$) and the complement of $F$ must be nonempty and have no greatest element. We must have $\gamma(t)>r$, and $\gamma(t')\leq r$ for all $t'<t$. Thus every rational in the nonempty real interval $(r,\gamma(t))$ is in the $\beta$-image of some elements of $L^*$, but none of these elements are in the $\gamma$-image of elements of $L$. What this means is that when constructing $L^*$ we inserted infinitely many elements below $t$ but above every element $t' < t$. This insertion could only have happened if $t$ covered an element of $L$. But if $t$ covers an element, then the complement of $F = [t,\infty)$ HAS a greatest element after all. This is a contradiction to the assumption that $F$ is not open, so the map $\gamma$ is continuous.

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  • $\begingroup$ This would have been my answer. But I would have shown that the subspace topology on $ L$ as a subspace of the ordered space $L^*$ co-incides with the order-induced topology on $L$ by considering the closure operation... Then, since $\beta:L^*\to \mathbb Q$ is an order-isomorphic homeomorphism, $\beta$ restricted to $L$ is an order-isomorphic homeomorphic embedding of the subspace $L$ into $\mathbb Q$.... BTW since $L^*$ \ $L$ is open in $L^*,$ you also have the result that $\gamma (L)$ is closed in $\mathbb Q .$ $\endgroup$ – DanielWainfleet May 13 '17 at 6:17

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