This question already has an answer here:

Prove that $\mathbb{R}_l$ is not a second countable. ($\mathbb{R}_l$ are the real ones with the topology of the lower limit)

I have tried to reason for the absurd and suppose that $\mathbb{R}_l$ has an countable basis but I can not find it, in that case any discrete subset of $\mathbb{R}_l$ would be countable, but I do not know how to find an countable base and if I find it I do not know how to find a discrete subset that is not countable.Could anyone help me, please? Thank you very much.


marked as duplicate by Community Nov 27 '17 at 0:31

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ Can you explain the notation? $\endgroup$ – ncmathsadist Nov 26 '17 at 23:45
  • $\begingroup$ @ncmathsadist Yes, I already edited it. $\endgroup$ – user402543 Nov 26 '17 at 23:47

You don't need to concern yourself with discrete subsets in this case.

For those unfamiliar with $\mathbb{R}_{l}$ this is the space whose underlying set is (obviously) $\mathbb{R}$ and whose topology is generated by the sets $[a,b)$ where $a,b\in\mathbb{R}$ $a<b$.

Assume that $\mathcal{B}$ is a countable basis for $\mathbb{R}_{l}$. Then, given a point $x$ let $B_{x}\in\mathcal{B}$ be a basis element such that $x\in B_{x}\subseteq[x,x+1)$. Then if $x\neq y$ we have that $B_{x}\neq B_{y}$. Why? Because $\inf B_{x}=x$ and $\inf B_{y}=y$. From this we may deduce that $\mathcal{B}$ must be uncountable as there is an injective function from $\mathbb{R}_{l}$ to $\mathcal{B}$ defined by $x\mapsto B_{x}$.

Note: This is as given in Munkres' section on separation axioms.

  • $\begingroup$ Oh, the half-open interval topology. $\endgroup$ – ncmathsadist Nov 27 '17 at 0:10
  • $\begingroup$ @ncmathsadist or traditionally “the Sorgenfrey line” $\mathbb{S}$. $\endgroup$ – Henno Brandsma Nov 27 '17 at 20:17

Not the answer you're looking for? Browse other questions tagged or ask your own question.