# Topology-Base for a topology

We know that the class of open intervals $(a,b)$, where $a,b$ are rational numbers is a countable base for $\mathbb R$.

But, $[a,b]$ where $a,b$ are rational numbers does not produce a base for $\mathbb R$.

Can we say that any $(a,b)$ or $[a,b]$ where $a$ is rational number and $b$ is an irrational number produce a base for $\mathbb R$?

• Obviously closed intervals can't, since they are not open in $\mathbb R$. For open intervals, remember that both the rationals and the irrationals are dense in $\mathbb R$. – Alex Becker Dec 24 '12 at 7:31
• Do you mean "base for the standard topology on the reals" or "base for some topology on the reals" ? – Henno Brandsma Dec 24 '12 at 8:31
• @Henno,Base for the standard topology on R. – ccc Dec 24 '12 at 8:42

I assume that when you say "a base for $\mathbb R$" you mean "a base for the standard topology on $\mathbb R$. With that, the answer to your question is no since $[a,b]$ is never an open set in the standard topology on $\mathbb R$.
If you also meant to ask whether the collection of all $(a,b)$ where $a$ is rational and $b$ is irrational forms a basis for the standard topology on $\mathbb R$ then the answer is yes.
I am not sure if this is related to your question at all, but we could say that the family $\{ [a,b]\ \colon a,b \in \mathbb R, a <c< b\}$ form a neighborhood basis at the point $c$ with respect to the Euclidean topology. This is saying that $c$ is in the interior of $U$ iff there is some closed interval of nonzero length such that $c \in [a,b] \subseteq U$. The notion of a neighborhood basis is different than a base for a topology but might lead to confusion.