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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 $\{[a,b]:a,b \in \mathbb{Q}\}$?

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    $\begingroup$ 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. $\endgroup$ Mar 6, 2020 at 10:27

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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.

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  • $\begingroup$ Does the set {[a,b]:a,b are rational and a<b}forms a base for any topology on R? $\endgroup$
    – Jasmine
    Mar 6, 2020 at 10:34
  • $\begingroup$ 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. $\endgroup$ Mar 6, 2020 at 10:44
  • $\begingroup$ As an exercise, for which topology on $\mathbb{R}$ is $A$ a base? $\endgroup$ Mar 6, 2020 at 10:45
  • $\begingroup$ Obviously if A forms a topology on R that will be different from usual topology. But which one it generates? $\endgroup$
    – Jasmine
    Mar 6, 2020 at 10:54
  • $\begingroup$ I once asked my professor that wether this set A forms a base for topology on R he said no but if the end points in A be like a is rational and b is irrational then the set A will be a bse for some topology $\endgroup$
    – Jasmine
    Mar 6, 2020 at 10:59
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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.

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  • $\begingroup$ If i use the concept that a collection of sets forms a base for usual topology on R if every open interval in R contains at least one basis element. How can i show that there does not exist any open interval in R which contain in [a,b] where a,b are rational numbers $\endgroup$
    – Stewen
    Mar 6, 2020 at 10:18
  • $\begingroup$ With your definition it is true that closed intervals with rational end points form a base. But this is not an accepted definition of a base. Do you have a reference where a base is defined this way? @Stewen $\endgroup$ Mar 6, 2020 at 10:22

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