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Let $X \subset \mathbb{R}$ and give $X$ its order topology. (When) is it true that $X$ is homeomorphic to some subspace of $Y \subset \mathbb{R}$?

Example: Let $X = [0,1) \cup \{2\} \cup (3,4]$, then $X$ is order isomorphic to $Y=[0,2]$ in a fairly obvious way, hence the order topologies on $X$ and $Y$ are homeomorphic. But, the order topology on $Y$ agrees with the subspace topology on $Y$, so this particular $X$ is indeed homeomorphic to a subspace of $\mathbb{R}$.

Example: Let $X$ equal the set of endpoints of the Cantor set. The order on $X$ is a bit strange (it's order isomorphic to $\mathbb{Q} \times \{0,1\}$ in the dictionary order with a largest and smallest element adjoined), so one might expect the order topology on $X$ to be strange, but... $X$ is clearly 2nd countable under the order topology (there are countably many points, hence countably many intervals) and any order topology is regular, so $X$ is metrizeable by Urysohn. It is also easy to see the order topology on $X$ has no isolated points so, by a theorem of Sierpinski (any countable metric space without isolated points is homeomorphic to $\mathbb{Q}$), the order topology on $X$ is homeomorphic to the subspace $Y = \mathbb{Q}$ of $\mathbb{R}$.

For any countable $X$, the argument above shows the order topology on $X$ is metrizeable. In fact, I think the order topology on any $X \subset \mathbb{R}$ is metrizeable by this theorem of Lusin (an order topology $X$ is metrizeable if and only if the diagonal in $X \times X$ is a $G_\delta$ set). But this doesn't preclude the possiblity of $X$ weird enough to not be homeomorphic to any subspace of $\mathbb{R}$. I'm not sure how one would arrange this though. If it can be arranged with $X$ countable there would need to be many isolated points to avoid Sierpinski's result for instance...

Advertisement: Brian's positive answer below actually proves something even better. Namely:

Theorem: Let $X \subset \mathbb{R}$. Then there is an order isomorphism (hence homeomorphism of order topologies) $\varphi$ from $X$ to a new set $Y \subset \mathbb{R}$ whose order topology and subspace topology coincide.

Roughly, approach is to take countable set $D \subset X$ such that each point in $X \setminus D$ is a 2-sided limit (topologically) of points in $D$. Using countability of $D$, one can find an order embedding $\varphi$ of $D$ into $\mathbb{Q}$ such that $\varphi(D)$ has no "topological gaps" that are not also "order gaps" (this is the property which makes the subspace and order topologies coincide). The final step is to extend $\varphi$ to the rest of $X$ and check nothing goes awry in doing so.

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note the even stronger theorem in my answer. –  Henno Brandsma Nov 23 '11 at 10:34

2 Answers 2

up vote 2 down vote accepted

This is a completely different argument from the previous one (and with a little luck a correct one this time), but the conclusion is the same: the answer is yes, always.

Suppose that $X\subseteq\mathbb{R}$. Let $\tau_o$ and $\tau_s$ respectively be the order and subspace topologies on $X$; certainly $\tau_o\subseteq\tau_s$. In the sequel all intervals will be taken in $X$ unless subscripted with $\mathbb{R}$, so that for example $$[x,\to)=\{y\in X:x\le y\},$$ and $$[x,\to)_\mathbb{R}=\{y\in\mathbb{R}:x\le y\}\;.$$

Let $N$ be the set of points of $X$ that have either an immediate predecessor or an immediate successor in $X$; clearly $N$ is countable.

Let $L=\{x\in X:[x,\to)\in\tau_s\setminus\tau_o\}$ and $R=\{x\in X:(\leftarrow,x]\in\tau_s\setminus\tau_o\}$; I’ll call points of $L$ left pseudogaps and points of $R$ right pseudogaps of $X$. It’s not hard to see that for any $x\in X$:

  • $x\in L$ iff there is $\epsilon>0$ such that $(x-\epsilon,x)_\mathbb{R}\cap X=\varnothing\ne(\leftarrow,x)$, and $x$ has no immediate predecessor in $X$; and
  • $x\in R$ iff there is $\epsilon>0$ such that $(x,x+\epsilon)_\mathbb{R}\cap X=\varnothing\ne(x,\to)$, and $x$ has no immediate successor in $X$.

It follows immediately that $L\cup R$ is countable. Moreover, it’s well-known that $\tau_s$ has a base of intervals, not necessarily open, in $X$, so $L\cup R$ is precisely the set of points at which $\tau_s$ and $\tau_o$ disagree.

Let $D_0$ be a countable $\tau_s$-dense subset of $X\setminus(L\cup R\cup N)$, and let $D=D_0\cup L\cup R\cup N$; we may assume that $D$ contains any endpoints of $X$. Enumerate $D=\{x_n:n\in\omega\}$ and $\mathbb{Q}=$ $\{q_n:n\in\omega\}$. Recursively define $\varphi:D\to\mathbb{Q}$ as follows: if $\varphi(x_k)$ has been defined for $k<n$, let $$m=\min\bigg\{i\in\omega:\forall k<n\big[q_i<\varphi(x_k)\leftrightarrow x_n<x_k\big]\bigg\}\;,\tag{1}$$ and set $\varphi(x_n)=q_m$. Clearly $\varphi$ is order-preserving.

Suppose that $x\in L$. Then $x=\sup_X(\leftarrow,x)\cap D$, and the construction guarantees that $\varphi(x)=$ $\sup_\mathbb{R}\varphi[(\leftarrow,x)\cap D]$. Similarly, $\varphi(x)=\inf_\mathbb{R}\varphi[(x,\to)]$ for each $x\in R$.

Now extend $\varphi$ to $X$ by setting $\varphi(x)=\sup_\mathbb{R}\{\varphi(y):y\in(\leftarrow,x)\cap D\}$ for $x\in X\setminus D\;$; clearly $\varphi$ is strictly order-preserving. Moreover, the construction ensures that if $x=\sup_X(\leftarrow,x)$, then $\varphi(x)=\sup_\mathbb{R}\varphi[(\leftarrow,x)]$, and if $x=\inf_X(x,\to)$, then $\varphi(x)=\inf_\mathbb{R}\varphi[(x,\to)]$. (This is a consequence of setting $\varphi(x_n)$ to $q_m$ with $m$ given by $(1)$.)

Let $Y=\varphi[X]$; clearly $Y$ with the order topology is homeomorphic to $\langle X,\tau_o\rangle$. Suppose that $y=\varphi(x)$ is a left pseudogap of $Y$. Then $\varphi(x)\ne\sup_\mathbb{R}\varphi[(\leftarrow,x)]$, so $x\ne\sup_X(\leftarrow,x)$. This is possible only if $x$ has an immediate predecessor $x^-$ in $X$, and in that case $(\varphi(x^-),y)=\varnothing$, and $y$ is not a left pseudogap of $Y$ after all. Similarly, $Y$ has no right pseudogaps, and therefore the subspace and order topologies on $Y$ coincide with each other, and $\langle X,\tau_o\rangle$ is homeomorphic to $Y$.

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Thanks for your reply, coincidentally, I was just about to post a proof for $X$ countable. A few minor comments, you should probably introduce a $d_L$ and a $d_R$ since I suppose that a point could be both a left and a right gap. Also I guess you mean "let $Z = \psi[Y]$" in your final paragraph? –  Mike F Nov 22 '11 at 20:00
    
Also I'm not sure I understand how this works, can't the function $\psi$ just collapse $X$ to a point? For example, take $X$ to be the endpoints of the Cantor set. It seems that for any $x_0$ in the middle somewhere and $x\in X$, $x < x_0$ we have $\sum_{y \in (x,x_0] \cap L} d(y)$ equal to the Lebesgue measure of the complement of $X$ in $[0,1] \cap (x,x_0]$, which would just be $x_0 - x$? In fact, I don't think it should be possible for any $Z$ to be order isomorphic to this $X$ and have neither left nor right gaps... –  Mike F Nov 22 '11 at 20:04
    
If $X$ is the ‘endpoints’ of the Cantor set, its order and subspace topologies are the same, so $L\cup R=\varnothing$, and no collapsing is needed. You’re right about $d_L$ and $d_R$ and about the typo at the end. –  Brian M. Scott Nov 22 '11 at 20:24
    
Ah, I see I was confused about your definition of gap. I will read through this again. –  Mike F Nov 22 '11 at 20:33
    
@Mike: However, the Cantor set minus the left endpoints of the deleted intervals does have a problem, so $\varphi$ can collapse points. If $x_0=2/3$, for instance, $\varphi(0)$ must clearly be $2/3$, by the argument that you just gave. –  Brian M. Scott Nov 22 '11 at 20:43

According to this paper, any second countable ordered space is (order) homeomorphic to a subspace of $\mathbb{R}$. Moreover, any separable one is homeomorphic to a subspace of the Sorgenfrey line. The proofs seem similar to Brian's.

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+1: Thank you, Henno. –  Mike F Nov 24 '11 at 7:40

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