# Equivalent definitions for GO-spaces (generalized ordered spaces)

GO-spaces (= generalized ordered spaces) are subspaces of LOTS (linearly ordered topological spaces). There are several definitions in use and I am wondering how to show the equivalence between them.

A subset $$A$$ of an ordered set $$(X,<)$$ is called order-convex if for each $$a,b\in A$$ with $$a, $$A$$ contains all points of $$X$$ between $$a$$ and $$b$$.

Proposition: Let $$X$$ be a topological space with topology $$\tau$$. The following are equivalent:

(a) There is a linear order $$<$$ on $$X$$ such that $$\tau$$ contains the order topology induced by the order $$<$$ and every point has a local base of $$\tau$$-open nbhds consisting of order-convex sets.

(b) $$X$$ is $$T_1$$ and there is a linear order $$<$$ on $$X$$ such that every point has a local base of $$\tau$$-open nbhds consisting of order-convex sets.

(c) $$X$$ is homeomorphic to a subspace of a LOTS.

Below is a proof that (a) and (b) are equivalent. Can someone provide a proof of the equivalence with (c)?

(a) implies (b):

Suppose $$<$$ is a linear order on $$X$$ as in condition (a). The order topology induced by $$<$$ is $$T_1$$ (actually even $$T_5$$ as for any LOTS). The topology $$\tau$$ is finer than that order topology, hence also $$T_1$$.

(b) implies (a):

Suppose (b) holds. The order topology induced by $$<$$ admits the collection of open rays $$(\leftarrow,x)$$ and $$(x,\to)$$ as a subbase. So to show $$\tau$$ is finer than that order topology, it is enough to show that these rays are $$\tau$$-open. Let $$x\in X$$. Since $$\{x\}$$ is closed, the set $$U=(\leftarrow,x)\cup(x,\to)$$ is open. Given $$w, there is some order-convex $$\tau$$-open nbhd $$V$$ of $$w$$ contained in $$U$$. But since $$V$$ is order-convex and does not contain $$x$$, it cannot contain points to the right of $$x$$. That is, $$V\subseteq(\leftarrow,x)$$, which shows that $$(\leftarrow,x)$$ is $$\tau$$-open. And similarly for $$(x,\to)$$.

Additional observation:

Given (a) or (b), every point $$x\in X$$ admits a local base of $$\tau$$-open nbhds of a special form.

Every open interval $$(a,b)$$ with $$a is $$\tau$$-open, since $$\tau$$ is finer than the order topology induced by $$<$$. Now, let $$V$$ be an order-convex nbhd of $$x$$. If $$V$$ contains points of $$X$$ both to the left and to the right of $$x$$, it necessarily contains an open interval $$(a,b)$$ around $$x$$. On the other hand, if $$V$$ contains no element of $$X$$ to the left of $$x$$, necessarily $$V\subseteq[x,\to)$$ and $$[x,\to)$$ is $$\tau$$-open (as the union of $$V$$ and $$(x,\to)$$). Similarly, if $$V$$ contains no element of $$X$$ to the right of $$x$$, necessarily $$V\subseteq(\leftarrow,x]$$ and $$(\leftarrow,x]$$ is $$\tau$$-open. Combining the various possibilities, it's easy to check that $$V$$ contains a $$\tau$$-open nbhd of $$x$$ having the form of an interval (open, half-open, or singleton interval) of the form $$(a,b)$$, $$[x,b)$$, $$(a,x]$$, or $$[x,x]=\{x\}$$ with $$a and/or $$x as the case may be. So the order-convex sets in a local base at $$x$$ can be chosen to be intervals of that form.

To summarize, every point of $$X$$ has a local base of $$\tau$$-open nbhds of (possibly degenerate) intervals of $$(X,<)$$.

It also follows that a subbase for the topology $$\tau$$ is given by the collection of sets consisting of:

• (1) all sets of the form $$(\leftarrow,x)$$ and $$(x,\to)$$ for $$x\in X$$;

• (2) all sets of the form $$(\leftarrow,x]$$ such that $$(\leftarrow,x]$$ is $$\tau$$-open, $$x$$ is not the maximum element of $$X$$ and $$x$$ has no immediate successor in $$(X,<)$$;

• (3) all sets of the form $$[x,\to)$$ such that $$[x,\to)$$ is $$\tau$$-open, $$x$$ is not the minimum element of $$X$$ and $$x$$ has no immediate predecessor in $$(X,<)$$.

Note that the sets in (1) generate the order topology on $$X$$ induced by $$<$$. The cases (1), (2), (3) above are mutually exclusive. A base for the topology is formed by finite intersections of sets in the subbase, which have the form of suitable open, half-open or closed intervals or rays in $$(X,<)$$.

Alternate characterizations:

Conditions (a) and (b) in the Proposition above were stated in terms of every point having a local base of $$\tau$$-open order-convex nbhds. But using the same kind of reasoning as in the "Additional observation" above, it's easy to see that it's enough to have a local base of order-convex nbhds without requiring them to be open for the topology.

(For example, as one of the cases to be considered, supposed $$V$$ is a (not necessarily open) order-convex nbhd of $$x$$ with $$V\subseteq[x,\to)$$ and $$V$$ contains an element $$b$$ with $$x. Then $$[x,\to)$$ is open in $$X$$, as the union of the open sets $$\operatorname{int}(V)$$ and $$(x,\to)$$. As $$(\leftarrow,b)$$ is also open in $$X$$, so is their intersection $$[x,b)$$, which is a $$\tau$$-open order-convex nbhd of $$x$$ contained in $$V$$. The other cases are handled similarly.)

To summarize:

In the Proposition above, conditions (a), (b), (c) are also equivalent to the following:

(d) There is a linear order $$<$$ on $$X$$ such that $$\tau$$ contains the order topology induced by the order $$<$$ and every point has a local base of order-convex nbhds.

(e) $$X$$ is $$T_1$$ and there is a linear order $$<$$ on $$X$$ such that every point has a local base of order-convex nbhds.

• Have you checked this paper by D. J. Lutzer ?
– Ulli
Commented May 16 at 13:10
• thanks, I'll take a look. Commented May 17 at 3:23

## 1 Answer

(c) implies (b):

Let $$Y$$ be a LOTS, with order $$<$$ and corresponding order topology $$\tau'$$. Let $$X$$ be a subset of $$Y$$, with subspace topology $$\tau$$ induced from $$\tau'$$, and with linear order equal to the restriction of the order on $$Y$$.

Since $$Y$$ is a LOTS, each point $$x\in X$$ has a local base of $$\tau'$$-open nbhds in $$Y$$ consisting of open intervals or open rays in $$(Y,<)$$, which are order-convex. Intersecting these with $$X$$ gives a local base of $$\tau$$-open nbhds of $$x$$ in $$X$$ that are order-convex in $$(X,<)$$. In addition, $$X$$ is $$T_1$$ as a subspace of a $$T_1$$ space.

(a) implies (c):

Suppose (a) holds; that is, $$(X,<)$$ is a linearly ordered set and $$\tau$$ is a topology on $$X$$ containing the order topology induced by $$<$$ and every point has a local base of $$\tau$$-open nbhds consisting of order-convex sets.

As explained in the "Additional obervation" section of the question, $$\tau$$ has a subbase consisting of the sets satisfying (1), (2), or (3). The sets in (1) form the usual subbase for the order topology on $$X$$ induced by $$<$$.

Now construct a larger ordered set $$(Y,<)$$ containing $$X$$ and extending the order on $$X$$ as follows. For each set $$(\leftarrow,x]$$ in (2), add an element $$x^+$$ as an immediate successor to $$x$$. And for each set $$[x,\to)$$ in (3), add an element $$x^-$$ as an immediate predecessor to $$x$$. Let $$\tau'$$ be the order topology on $$Y$$ induced by this order, with makes $$Y$$ into a LOTS.

Claim: The subspace topology on $$X$$ induced by $$\tau'$$ coincides with $$\tau$$.

For convenience, let's write intervals in $$Y$$ with a subscript $$Y$$ and intervals in $$X$$ without subscript. To show the claim, take the subbase for $$(Y,\tau')$$ consisting of all intervals $$(\leftarrow,y)_Y$$ and $$(y,\to)_Y$$ with $$y\in Y$$ and look at the intersection of each of them with $$X$$. The different cases are, with $$x\in X$$ and $$x^+$$ and $$x^-$$ as defined above:

• $$X\cap(\leftarrow,x)_Y=(\leftarrow,x)$$ and $$X\cap(x,\to)_Y=(x,\to)$$;
• $$X\cap(\leftarrow,x^+)_Y=(\leftarrow,x]$$ and $$X\cap(x^+,\to)_Y=(x,\to)$$;
• $$X\cap(\leftarrow,x^-)_Y=(\leftarrow,x)$$ and $$X\cap(x^-,\to)_Y=[x,\to)$$.

The subbase obtained for the subspace topology consists of all the sets satisfying conditions (1), (2), (3), which form a subbase for the topology $$\tau$$.

Notes:

1. It is easy to check that $$X$$ is a topologically dense subset of the LOTS $$Y$$. Let $$(Z,<)$$ be the Dedekind-MacNeille completion of the ordered space $$(Y,<)$$. The ordered set $$Z$$ is order-complete, so it is compact with its order topology. It is a general fact that $$Y$$ is topologically dense in $$Z$$ with its order topology; so $$X$$ is also topologically dense in $$Z$$. Thus every GO-space is homeomorphic to a dense subspace of a compact LOTS (such that the homeomorphism preserves the order).

2. The construction in the paper

Lutzer, David J., On generalized ordered spaces, Dissertationes Math., Warszawa 89, 32 p. (1971). ZBL0228.54026.

is a little different. For case (2), instead of a successor $$x^+$$ after $$x$$, it adds a chain of successors, i.e., a copy of $$\omega=\{1<2<3<\dots\}$$. For case (3), instead of adding a predecessor $$x^-$$ before $$x$$, it adds a chain of predecessors, i.e., a copy of $$\omega^*=\{\dots<-3<-2<-1\}$$. The resulting ordered set $$(X^*,<)$$ gives a LOTS that contains $$X$$ as a closed subspace.