Show that every order topology is regular

Question

If someone could review my proof below for correctness that would be appreciated. Thanks!

Problem:

Show that every order topology is regular.

Proof:

Let X be a topological space with the order topology.

One point sets are closed since given {x} we have that for any point y $\gt$ x and y $\in$ X - {x} we have the open set (x,y$_1$) where y$_1$ $\gt$ y. Hence y is not a limit point of {x} since (x, y$_1$) does not intersect {x}. The case for y $\lt$ x follows similarly. If y is the smallest or largest element in X then the open sets (x,y] or [y,x) are open sets containing y that do not intersect {x}. Hence {x} is closed.

Now consider the following theorem:

X is regular if and only if given a point x of X and a neighborhood U of x, there is a neighborhood V of x such that $\bar{V}$ $\subset$ U.

Now let x $\in$ U for U open in X. Then some basis element U$_b$ contains x for U$_b$ $\subset$ U. If U$_b$ = {x} then U$_b$ = V is a neighborhood of X such that $\bar{V}$ $\subset$ U since U$_b$ is a closed set and U$_b$ $\subset$ U.

So assume U$_b$ has more than one point. Suppose U$_b$ is of the form (c,...y,...x,..,d) where U$_b$ has at least 2 points. If there are two points y and z such that c < y < x and x < z < d then the open set V = (y,z) is an open neighborhood containing x such that $\bar{V}$ $\subset$ U$_b$ since for any n $\gt$ z the open set (z,n$_1$) for any n$_1$ $\gt$ n contains n and does not intersect V. Hence no n $\gt$ z can be a limit point of V. A similar argument extends to any n $\lt$ y. Hence $\bar{V}$ $\subset$ U$_b$. Note that where n is a max or min point of X the boundary open sets (z,n] or [n,y) suffice in the above argument.

Note that either y or z must be a member of U$_b$ = (c,d) since otherwise (c,d) is a single point set only containing x. So assume only y is a member of (c,d). Then the set (y, d) = V, for y $\lt$ x, satisfies as an open neighborhood containing x such that $\bar{V}$ $\subset$ U$_b$. For by a similar argument to above any n $\lt$ y is not a limit point of V and d is also not a limit point of V since the open set (x,d$_1$) or (x,d$_1$], for d$_1$ $\gt$ d, are open sets that contain n and do not intersect V since there is no point in the interval (x,d). So $\bar{V}$ $\subset$ U$_b$. Note a similar argument applied if we assume z $\in$ (c,d) instead of y.

If U$_b$ were instead of the form (c,d] or [d,c) the argument is similar with an important added detail that to show points outside of U are not limit points of V you can use the boundary open sets (z,d] or [c,y). I've omitted other details for brevity.

Hence since V $\subset$ U$_b$ $\subset$ U so V satisfies the theorem posted above and hence X is regular.

Answer 1

Alternative proof without the awkwardness of limit points.

Assume x in open U.
Thus exists a,b with x in (a,b) subset U.

If exists u,v with a < u < x < v < b, then
x in open (u,v) and $\overline{(u,v)} \subseteq [u,v] \subseteq U.$

If (a,x) is empty and exists v with x < v < b, then
x in open (a,v) = [x,v) and $\overline{[x,v)} \subseteq [x,v] \subseteq U.$

If (x,b) is empty and exists u with a < u < x a simular proof ensues.

If (a,x) and (x,b) are empty, then
(a,b) = {x} is open and as linear ordered
topologies are $T_1,$ {x} is closed subset of (a,b).

Answer 2

The answer by @William Elliot is not a general proof. In fact, if we take the ordered space $X = [0,1], x = 1$ and $U = (0.5,1]$, then there are no $a,b$ such that $x \in (a,b) \subset U$. Thus we have to consider two more cases. The first case is: there is $a \in X$ such that $x \in (a, \rightarrow) \subset U$. In this case, if there is $c \in X$ such that $a < c < x$ then $x \in (c,\rightarrow) \subset \overline{(c,\rightarrow)} \subset U$. If there is no such $c$, then $\{x\}$ is open as well as closed and hence $x \in \{x\} \subset \overline{\{x\}} \subset U$. Similarly the second case $x \in (\leftarrow, a) \subset U$ can be handled.