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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 $\mathbb R$?

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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
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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

2 Answers 2

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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.

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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.

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