Sorgenfrey line topology

For every $x\in\mathbb R$ we have $\mathcal{B}(x):=\{[x,z): z>x\}$ as the sorgenfrey line.

First I want to show that there is topology $\tau$ defined on $\mathbb R$, i.e I have to show that $\mathcal{B}(x)$ satifies the properties of a neighborhood basis. I succeded in showing two parts, the last one $\forall V\in\mathcal{B}(x)\exists V_0\in\mathcal{B}(x)\forall z\in V_0\exists W\in\mathcal{B}(z):W\subseteq V$ is still remaining. Graphically its clear, but I do not know how to write it down formally.

Secondly I have a non-empty bounded interval I, such that $(a,b)\subseteq I\subseteq [a,b]$ for $a,b\in\mathbb R$ with $a<b$. I want to show that $I$ is open with repsect to the topolgy $\tau$ $<=> b\notin I$

My idea: I already shows that if $(X,\tau)$ is a topologocal space and for every $x\in X$ there is a neighborhood basis $\mathcal{B}(x)$, then for $I\subseteq X$: $I$ open$<=>\forall x\in I\exists V\in \mathcal{B}(x): V\subseteq I$

May you have an idea how to use this (I think it should work with this lemma) to prove the equivalence.

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What $V_0$ do you have in mind for the last property of neighborhood base? Also you should change the $x$ to $z$ at the end of the formula. – Stefan Hamcke Apr 15 '13 at 13:42
I guess an interval like $V_0=[x,a)$ and $V=[x,b)$ where $a\le b <x$ – Babla Apr 15 '13 at 13:45
for simplicity you could take $V_0=V=[x,b)$ itself. Then for $z\in V$ what is the simplest interval $[z,a)$ contained in $V$?. – Stefan Hamcke Apr 15 '13 at 13:48

For that last property it suffices to show is that if $y\in V\in\mathscr{B}(x)$, then there is a $W\in\mathscr{B}(y)$ such that $W\subseteq V$. Since $V\in\mathscr{B}(x)$, $V=[x,z)$ for some $z>x$, and since $y\in V$, $y<z$; now let $W=[y,z)$ and verify that it has the required properties.
For the second part, if $b\notin I$, what is $\bigcup_{x\in I}[x,b)$? And if $b\in I$, is there any $V\in\mathscr{B}(b)$ such that $V\subseteq I$?
He has stated it correctly, but he has to replace the last $x$ by $z$. – Stefan Hamcke Apr 15 '13 at 13:52
I think than $I=(a,b)$, correct?, if $b\in I$ than there cant be a $V\subseteq I$ – Babla Apr 15 '13 at 13:54
@Babla: $\bigcup_{x\in I}[x,b)$ is simply $I$ if $b\notin I$; this could be $(a,b)$, but it could also be $[a,b)$. You’re right about the other direction: there is no $V\in\mathscr{B}(b)$ with $V\subseteq I$. – Brian M. Scott Apr 15 '13 at 13:57
@Stefan: The $V_0$ in his version is completely superfluous: it does nothing. – Brian M. Scott Apr 15 '13 at 13:58
So this is it? I open with repsect to $\tau$ simply means that $b\in I$?. I have one more question. Which of the $\tau$ open intervals are also $\tau$ closed? – Babla Apr 15 '13 at 13:58