Open sets which are not closed in the Sorgenfrey line Basically, it is a simple fact about the Sorgenfrey line that:


*

*the only connected sets are the singelton sets.

*the open set in Sorgenfrey line $(b,\infty)$ is not closed.
But are there other open sets which not closed?
The argument for 1 and 2 are not difficult. Do you think I am right?
Any help will be appreciated.
 A: Let $A$ be a subspace of the Sorgenfrey line $S$ and let $a,b\in A$ with $a<b.$ Then $[b,\infty)$ and $(-\infty,b)=\cup_{x<b}[x,b)$ are open in $S.$ So $A\cap (-\infty,b)$ and $A\cap [b.\infty)$ are relatively open in $A,$ are disjoint, and are not empty, and their union is $ A.$ So $A$ is not connected. So the only connected subspaces of $S$ are the empty set and the $1$-element subsets.
Any open $B\subset S$ such that $(b,c)\subset B$ and $b\not \in B$ is not closed, because any nbhd $U$ of $b$ covers $[b,d)$ for some $d>b,$ so $U\cap B\supset (b,\min (c,d))\ne \emptyset.$ So $b\in \bar B$ \ $B.$ This however is not necessary  in order that $B$ be open but not closed. For example if $B=\cup_{n\in N} [\frac {1}{2n},\frac {1}{2n-1})$ then $0\in \bar B$ \ $B.$ An open $B\subset S$ is not closed iff there exists $b\in S$ \ $B$ such that $b=\inf\; ((b,\infty)\cap B\;).$
A: To show that the singletons are the only connected sets you need to argue like this:
Let us consider the set $\{a,b\}$ where say $a<b$.
Then the sets $(-\infty ,\dfrac{a+b}{2})\cup [b,\infty 
)$ are both open sets in the sorgenfrey line which form a disconnection of $\{a,b\}$.
