# Base for usual topology on $\mathbb{R}$

I know that the set $$\{(a,b):a,b \in \mathbb{Q}\}$$ forms a base for usual topology on $$\mathbb{R}$$ but why not this set $$\{[a,b]:a,b \in \mathbb{Q}\}$$?

• It is a so-called network for the topology, provided we always have $a<b$ in these intervals. A network is like a base * without * the requirement that all members are open. Mar 6, 2020 at 10:27

I you took $$\{[a,b] : a,b \in \mathbb{Q}\}$$ to be the basis of your topology, then in particular all of those sets would be open. So for example $$[0,1]$$ would be open, which is not the case in the standard topology.
• Yes, it does. One needs to check that $A:=\{[a,b] : a,b \in \mathbb{Q}, a < b\}$ covers $\mathbb{R}$ and that $\forall [a,b], [c,d'] \in A$ and $x \in [a,b] \cap [c,d]$ there exists $[e,f] \in A$ s.t. $x \in [e,f] \subseteq [a,b] \cap [c,d]$; and this is the case here. Mar 6, 2020 at 10:44
• As an exercise, for which topology on $\mathbb{R}$ is $A$ a base? Mar 6, 2020 at 10:45
A base for a topology is a subcollection of the topology. Since $$[a,b]$$ is not open in the usual topology these sets do not form a base.