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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 $\mathbb{R}$?

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up vote 3 down vote accepted

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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In [0,1]∩[1,2],why can't I find any two rational numbers to satisfy the 2nd condition? –  ccc Dec 17 '12 at 5:05
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@ccc: As I mentioned, this point depends on whether you allow degenerate intervals or not. If you do not, then note that $[ 0 , 1 ] \cap [ 1 , 2 ] = \{ 1 \}$, and so the only set satisfying $1 \in V \subseteq [ 0 , 1 ] \cap [ 1 , 2 ]$ is $V = \{ 1 \} = [1,1]$, which (under the current discussion) is not allowed. If you do allow such intervals, there is no problem. –  Arthur Fischer Dec 17 '12 at 5:12
    
@ Arthur,Yeah,I'm not talking about degenerate intervals.In the 2nd condition,given U_1,U_2 in B means that U_1 and U_2 must be closed intevals with rational end points.Isn't it?So can I take [0,1] and [1,2] as closed intervals with rational end points? –  ccc Dec 17 '12 at 5:24
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@ccc: Aren't $0$, $1$ and $2$ rational? –  Arthur Fischer Dec 17 '12 at 5:25
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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.

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