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I can prove that $(a,b)$ where $a,b$ are rational numbers form a countable base for the topology on $\mathbb{R}$.

But, how to show that the collection $[a,b]$ where $a$ and $b$ are rational numbers, is not a base for a topology on $\mathbb{R}$?

2 Answers 2

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Note that to be a base for a topology on a set $X$, a collection $\mathcal{B}$ need only satisfy the following two conditions:

  1. $\bigcup \mathcal{B} = X$; and
  2. given $U_1 , U_2 \in \mathcal{B}$ and $x \in U_1 \cap U_2$ there is a $V \in \mathcal{B}$ such that $x \in V \subseteq U_1 \cap U_2$.

Now the collection $\mathcal{B}$ of closed intervals in $\mathbb{R}$ with rational endpoints clearly satisfies the first condition. As for the second condition, it depends on the details. If you allow degenerate intervals such as $[a,a] = \{ a \}$, then $\mathcal{B}$ will satisfy the second condition. If not, then it won't: consider $[ 0 , 1 ] \cap [ 1 , 2 ]$.

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    @ccc: Aren't $0$, $1$ and $2$ rational?2012-12-17
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The collection you mention is a base for a topology on $\mathbb R$. It is a base for a topology different than the standard topology on $\mathbb R$ for which you described a countable basis. Since $[a,b]$ is not open in the standard topology these can't form a base for the standard topology.