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

Asked Sep 18 '18 at 14:27

Active Sep 18 '18 at 14:27

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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?

asked Sep 18 '18 at 14:27

Math geek

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
For the usual topology, it is not true that, given a basis $\mathcal B$, for each $x\in\mathbb R$ there is a $B_x\in\mathcal B$ such that $\min B_x=x$. Actually, there is no open set $A$ such that $x\in O$ and that $\min A=x$. – José Carlos Santos Sep 18 '18 at 14:52
I have still doubt on why minimum is taking? – Math geek Sep 18 '18 at 15:03
Can't we take infimum for usual topology case ? – Math geek Sep 18 '18 at 15:05
Sure we can. But there's reason why the function will be one-to one. And I shall answer no more questions here. If you have a new question, post it as such. – José Carlos Santos Sep 18 '18 at 15:07

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

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