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

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?

• Isn't he proved that there exist an uncountable basis in the proof? – Math geek Sep 18 '18 at 14:40
• The first answer is fine. – José Carlos Santos Sep 18 '18 at 14:41
• How does in the proof guarentee there doesnot exist a countable basis? – Math geek Sep 18 '18 at 14:42
• 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 $\mathbb R$. – José Carlos Santos Sep 18 '18 at 14:44
• Where will the proof fail,if we consider usual topology on R? – Math geek Sep 18 '18 at 14:49