Let $\tau$ be lower limit topology on $\mathbb R.$ Then $(X,\tau)$ is not second countable.

Suppose on the contrary there exists a countable basis for $(X,\tau)$. Let $\mathscr B=\{B_1,...,\}$ be the basis. Any non-empty open set is the union of elements of $\mathscr B.$ Where will I get a contradiction? How do I proceed the proof?

  • $\begingroup$ Isn't he proved that there exist an uncountable basis in the proof? $\endgroup$ – Math geek Sep 18 '18 at 14:40
  • $\begingroup$ The first answer is fine. $\endgroup$ – José Carlos Santos Sep 18 '18 at 14:41
  • $\begingroup$ How does in the proof guarentee there doesnot exist a countable basis? $\endgroup$ – Math geek Sep 18 '18 at 14:42
  • $\begingroup$ The author proved that, given a basis $\mathcal B$, there is a one-to-one function from $\mathbb R$ into $\mathcal B$. Therefore, $\mathcal B$ has at least the cardinal of $\mathbb R$. $\endgroup$ – José Carlos Santos Sep 18 '18 at 14:44
  • $\begingroup$ Where will the proof fail,if we consider usual topology on R? $\endgroup$ – Math geek Sep 18 '18 at 14:49

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