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I am absolutely clueless here, I know that $\mathbb{R}$ has a countable dense subset, if below propositions hold true:

The rationals $\mathbb{Q}$ are a countable set, and when you give $\mathbb{R}$ the standard (order) topology, then the rationals $\mathbb{Q}$ are dense in $\mathbb{R}$.

So, based on these definitions, if I declare $[0,1)$ and $[1,2)$ which follows the notaions of floor topology, then how should I prove or show that they have a countable basis and dense subset in $T_{ll}$?

Need help on this.

Math_Is_Fun
  • 692
  • $\Bbb Q$ is still dense. It intersects every base element of the standard base. – Henno Brandsma Mar 08 '20 at 07:58
  • How to solve problems 3.68 and 69? I know the definitions of standard order topology and floor topology but I really don't know how to make use of it. – Math_Is_Fun Mar 08 '20 at 08:01
  • Have you looked at the duplicate? I gave you the argument for 3.68 already in the other comment: a set is dense iff it intersects every non-empty base element. $(a,b) \subseteq [a,b)$ and you already know $\Bbb Q$ is dense in the standard topology! – Henno Brandsma Mar 08 '20 at 08:03

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