A topology on a dense poset, equal to order topology if linear, coarser otherwise

Question

I am looking for a topology on a dense partially ordered set $(X,\leq)$, such that this topology equals the order topology in case $(X,\leq)$ is linear, and in which any open set $U$ satisfies

1. for each non-minimal $x$ in $U$ and any $v < x$ there is an element $w\in[v,x)$ such that $(w,x]\subseteq U$, and
2. for each non-maximal $x$ in $U$ and any $z > x$ there is an element $y\in(x,z]$ such that $[x,y)\subseteq U$.

(Here $[v,x)=\{u:v\leq u<x\}$ and $(w,x]=\{u:w<u\leq x\}$ for any $v,w,x\in X$.)

In some way, as illustrated in the example below, this topology would closer resemble the Euclidean topology than the order topology would in case $X$ is a subset of $\mathbb{R}^n$, $n\geq 2$.

My questions are: Are there any common topologies in the literature with the above properties? And is the collection of sets satisfying conditions 1 and 2 a topology?

The order topology, generated by the sets $\{x:x<a\}$ and $\{x:x>a\}$, $a\in X$, satisfies the conditions 1 and 2 if $(X,\leq)$ is linear, but otherwise it may be too fine to satisfy them.

I have tried taking conditions 1 and 2 as defining conditions. If $\tau$ is the collection of all sets $U$ satisfying conditions 1 and 2, then it is not difficult to show that $\tau$ contains the empty set and $X$, and that if $\gamma$ is any subcollection of $\tau$ then the union of $\gamma$ lies in $\tau$. I have not yet managed to verify whether the intersection of any two elements in $\tau$ is an element in $\tau$.

The following example may help to illustrate things. Suppose that $X$ is the set $\{(x,y)\in\mathbb{R}^2:x=0\text{ or }y=0\}$, and let $\leq$ be a partial order on $X$ defined by \begin{equation*} (x_1,y_1)\leq(x_2,y_2) \Leftrightarrow \left\{\begin{array}{rl} x_1<0\text{ and }x_1\leq x_2,&\text{or}\\ x_1>0\text{ and }x_1\geq x_2,&\text{or}\\ x_1=x_2=0\text{ and }y_1\leq y_2. \end{array}\right. \end{equation*} In the order topology a set $U=\{(0,y):-1<y<1\}$ would be open, and the conditions 1 and 2 are not satisfied, because for any $\delta>0$ the intervals $((-\delta,0),(0,0)]$ and $[(0,0),(\delta,0))$ are no subsets of $U$. I would like to have a topology on $X$ such that any open set $U$ containing $(0,0)$ will contain a subset \begin{equation*} \{(x,y):-\delta<x<\delta\text{ and }y=0,\text{ or }x=0\text{ and }-\delta<y<\delta\} \end{equation*} for some $\delta>0$ (an "open ball" of radius $\delta$). It is like the Euclidean topology in $\mathbb{R}^2$, but restricted to $X$.

Answer 1

There has been much research about topologizing (partially) ordered sets; none of it is conclusive, satisfying or simple. Included is some stuff about linear order and topology that may interest you. If you do not web find any reaseach papers about order and topology, I'll look into my paper files and show what I've found.

strong order topology for ordered S is the topology generated by
{ (x,->), (<-,x), S - {x} : x in S }

T1 topology for ordered S
S_b = { x | not x <= b } = S\down b
S^b = { x | not b <= x } = S\up b
subbase for S = { S_b, S^b | b in S }

Order and Topology, at.yorku.ca/t/a/i/c/05.htm.
An ordered topological space is a topological space (X,T) equipped with a partial order <=. Usual compatibility conditions between the topology and order include convexity (T has a basis of order-convex sets) and the T_2-ordered property: <=, ie { (x,y) | x <= y }, is closed in XxX.

Since every topological space (X,T) can be considered as a trivially ordered space (X,T,=), the theory of ordered spaces includes the usual topological theory as a special case. Other important cases, each with its own techniques, include the totally ordered spaces and lattices.

-- at.yorku.ca/t/a/i/c/36.htm
A generalized ordered space (GO space) is a Hausdorff space X
with a linear ordering such that the topology has an open base
consisting of order convex sets.

If the topology coincides with the open interval topology of the order, then X is a linearly ordered topological space (LOTS). It is known that the class of GO spaces is topologically the same as the class of subspaces of LOTS. Thus GO spaces are monotonically normal.

Sorgenfrey line is GO space

--
Chris Good http://web.mat.bham.ac.uk/C.Good/research/pdfs/nests.pdf
mentions the van Mill & Wattel characterization of ordered spaces:

A space X is orderable iff there are 2 interlocking nests L and R of subsets of X, that together form a T_1-separating subbase for X. For definitions, see Good's paper; if we remove "interlocking" we get exactly a GO-space.

Answer 2

The collection $\tau_i$ of sets $U$ satisfying condition $i=1$ or $2$ defines a topology on $X$. (It then follows that the collection $\tau=\tau_i\cap\tau_2$ of sets satisfying both conditions 1 and 2 is a topology on $X$.)

A proof that $\tau_1$ defines a topology on $X$ is as follows.

The empty set and $X$ both satisfy condition 1.

Suppose that $\gamma$ is a collection of sets satisfying condition 1. If $x$ is a non-minimal element in $\bigcup\gamma$, then it is an element in $U$ for some $U$ in $\gamma$, and for any $v < x$ there is an element $w\in[v,x)$ such that $(w,x]\subseteq U\subseteq\bigcup\gamma$. So $\bigcup\gamma$ satisfies condition 1.

Suppose that $U_1$ and $U_2$ are any two elements in $\tau_1$. If $U_1\cap U_2$ is empty, then it lies in $\tau_1$. Suppose that $U_1\cap U_2$ is not empty. Let $x$ be any element in $U_1\cap U_2$. If $x$ is non-minimal, then for any $v < x$ there is an element $w_1$ in $[v,x)$ such that $(w_1,x]\subseteq U_1$. But since $w_1 < x$, there is an element $w_{12}$ in $[w_1,x)$ such that $(w_{12},x]\subseteq U_2$. Hence $(w_{12},x]\subseteq (w_1,x]\cap U_2\subseteq U_1\cap U_2$. So $U_1\cap U_2$ satisfies condition 1.