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 $a<b$, $A$ 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 $w<x$, 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 $a<x<b$ 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 $a<x$ and/or $x<b$ 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 $x<b$. 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.
