# Lower limit topology closed disjoint sets

I'm currenctly studying topology and I was wondering if you could help me with the following:

$X \subset \mathbb{R}$

Prove that if $A, B \subset \mathbb{R}$ are closed in $\mathcal{T}$ which is generated by $\mathcal{B} = \{ [a,b) \ | \ a,b \in \mathbb{R}, \ a<b \}$

and $A \cap B = \emptyset$, then $\exists U, V \in \mathcal{T} : \ A \subset U, \ B \subset V, \ U \cap V = \emptyset$

Could you help me with that?

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Hint: For $x \in A \subseteq \mathbb R \setminus B$ there is an $\epsilon_x > 0$ such that $[ x , x + \epsilon_x ) \subseteq \mathbb R \setminus B$. Similarly for $y \in B$ there is a $\delta_y > 0$ such that $[ y , y + \delta_y ) \subseteq \mathbb R \setminus A$. Can $[ x , x + \epsilon_x ) \cap [ y , y + \delta_y ) \neq \emptyset$ for $x \in A$ and $y \in B$?
Well. From what you've written, this intersection has to be empty, because $[ x , x + \epsilon_x ) \subset \mathbb{R} \setminus B$ and $[ y , y + \delta_y ) \subset \mathbb{R} \setminus A$. I can see in the picture that the first interval must have one endpoint in $A$ and $x + \epsilon_x$ can belong to $X \setminus A$ but it cannot be a member of $B$, and for the second interval accordingly, so there is no way they can intersect, but I am not sure how to prove that a bit more formally. – Hagrid Mar 2 '13 at 7:35
@Hagrid: If $[ a , b ) \cap [ c , d ) \neq \emptyset$ we can show that $\max \{ a , c \} \in [ a , b ) \cap [ c , d )$. The observations in your comment should allow you to derive a contradiction. – arjafi Mar 2 '13 at 8:00
So if $[ x , x + \epsilon_x ) \cap [ y , y + \delta_y ) \neq \emptyset$, then $\max \{ x , y \} \in \mathbb{R} \setminus (A \cup B)$ and this is a contradiction. Is that right? – Hagrid Mar 2 '13 at 8:06
@Hagrid: Yes! since that lower end-point must belong to either $A$ or $B$. – arjafi Mar 2 '13 at 8:07