A second countable GO-space can be homeomorphically embedded in a second countable LOTS

Question

A GO-space (generalized ordered space) is a topological space $X$ with topology $\tau$ together with a linear order $<$ such that $X$ is $T_1$ and every point has a local base of $\tau$-open neighborhoods consisting of order-convex sets. Examples are the Sorgenfrey line, the Michael line, and every linearly ordered topological space (LOTS).

As discussed for example in Equivalent definitions for GO-spaces (generalized ordered spaces), there are various equivalent characterizations. In particular, GO-spaces are the spaces homeomorphic to a subspace of a LOTS.

FACT: If the GO-space is second countable, the LOTS it embeds into can be chosen second countable.

This is mentioned in Remark (5.7) in this paper, but as a consequence of some non-trivial results about metrizability and other things. I feel like there should be a more direct proof of this. Can anyone provide one?

NOTE: This fact together with Every second countable LOTS is embeddable in $\mathbb R$ has an interesting consequence:

Every second countable GO-space can be embedded in $\mathbb R$.

And the proofs show that the homeomorphic embeddings into a second countable LOTS and into $\mathbb R$ can be chosen to be order-preserving.

Answer 1

Suppose $X$ is a GO-space with a linear order $<$ and a topology $\tau$ containing the order topology induced by $<$ and such that every point has a local base consisting of order-convex open nbhds. Assume $\tau$ is second countable.

As explained in Equivalent definitions for GO-spaces (generalized ordered spaces), one can embed $X$ into a specific LOTS $Y$ in an order-preserving fashion such that $\tau$ coincides with the subspace topology induced from $Y$. I'll show that $Y$ is also second countable.

$Y$ is constructed from $X$ by adding some points:

• (1) for each $x\in X$ such that $[x,\to)$ is $\tau$-open, $x$ is not the minimum element of $X$ and $x$ has no immediate predecessor in $(X,<)$, add an element $x^-$ as an immediate predecessor to $x$.
• (2) for each $x\in X$ such that $(\leftarrow,x]$ is $\tau$-open, $x$ is not the maximum element of $X$ and $x$ has no immediate successor in $(X,<)$, add an element $x^+$ as an immediate successor to $x$.

It is enough to show that $Y$ has a countable subbase (the collection of finite intersections of elements of the subbase will then form a countable base). A subbase for the topology of $Y$ consists of all intervals $(\leftarrow,y)_Y$ and $(y,\to)_Y$ for $y\in Y$, where the subscript $Y$ indicates an interval in $(Y,<)$. We'll build a countable subbase as a suitable subcollection of that first subbase. Since the situation is symmetric, we can focus attention on the intervals $(y,\to)_Y$ unbounded to the right.

(Fact *) By second countability of $X$, there are only countably many $x\in X$ such that $[x\to)$ is $\tau$-open in $X$ (reason: if $\mathscr B$ is a countable base for $\tau$, there is some $B_x\in\mathscr B$ and $x$ can be recovered from $B_x$ as its minimum element). And similarly for those $x\in X$ such that $(\leftarrow, x]$ is $\tau$-open in $X$.

In particular, only countably many elements $x^-$ and $x^+$ were added.

Also by second countability, $X$ is separable; so there is a countable subset $A\subseteq X$ that is $\tau$-dense in $X$.

Now consider various cases for $y$.

If $y=x^-$ or $y=x^+$, there are only countably many possibilities. Add the corresponding $(y,\to)_Y$ to the subbase.

Suppose $y=x\in X$.

If $(\leftarrow,x]$ is $\tau$-open in $X$, there are only countably many possibilities by (Fact *). Add $(x,\to)_Y$ to the subbase.

Otherwise, $(\leftarrow,x]$ is not $\tau$-open, so $x$ has no immediate successor in $(X,<)$ and $x$ is not the maximum element of $X$. For each $z\in X$ with $x<z$, there is some $a\in A$ with $x<a<z$ by density of $A$ in $X$. Also in this case, no point $x^+$ was added in the construction of $Y$. So for every point $z^-$ or $z^+$ in $Y\setminus X$ with $x<z^-$ or $x<z^+$ as the case may be, there is also an $a\in A$ between $x$ and such an element. This shows that the interval $(x,\to)_Y$ in $(Y,<)$ is the union of intervals of the form $(a,\to)_Y$ with $a\in A$. And we can take the countably many such $(a,\to)_Y$ to be part of the new countable subbase.

Now repeat for intervals unbounded to the left and we have constructed a countable subbase for $Y$.

Answer 2

Posting an incomplete answer as I've run out of time to work on it, but hope it is helpful. (Others are invited to fix it if they choose, or point out where it may be flawed.)

First, choose a linear order $<$ on $X$ such that its topology contains the order topology induced by the order $<$, and every point has a local base consisting of order-convex sets. By second-countability, we may choose a countable basis $\mathcal B=\{B_n:n<\omega\}$ of order-convex sets.

We extend $X$ as follows. Let $X_0=X$, and suppose $X_n\supseteq X$ has been defined and has an order $<$ compatible with $X$'s original order. Consider the order-convex subset $B_n\subseteq X$. We will define $a_n,b_n$ such that $B_n=(a_n,b_n)\cap X$. If $B_n$ has an infimum $a\in X_n\setminus B_n$, let $a_n=a$. If not, then we create a new point $a_n\in X_{n+1}\setminus X_n$ such that $x<B_n\Leftrightarrow x<a_n$ for all $x\in X_n$. If $B_n$ has a supremum $b\in X_n\setminus B_n$, let $b_n=b$. If not, then we create a new point $b_n\in X_{n+1}\setminus X_n$ such that $x>B_n\Leftrightarrow x>b_n$ for all $x\in X_n$.

It is immediate that $X$'s topology coincides with its subspace topology from $X_\omega=\bigcup\{X_n:n<\omega\}$, where $X_\omega$ is given its order topology. To see that $X_\omega$ is second-countable, I will show that $\{(\leftarrow,b_n):n<\omega\}\cup\{(a_n,\rightarrow),n<\omega\}$ is a countable subbasis. Let $x\in(y,\rightarrow)$; we will find $n<\omega$ with $x\in (a_n,\rightarrow)\subseteq (y,\rightarrow)$ (and omit the corresponding argument for $x\in(\leftarrow,y)$).

If $x\in X$, we have that $(y,\rightarrow)\cap X$ is an open neighborhood of $x$, and we may choose $n<\omega$ with $x\in B_n\subseteq (y,\rightarrow)\cap X$. I claim $a_n\leq y$. (TBC)