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

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 $\mathbb R$?

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

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.

share|cite|improve this answer

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.

share|cite|improve this answer

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.