Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 2 down vote accepted

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 $V\subseteq I$?

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.