I am trying to prove that $\mathbb{R}$ with the lower limit topology is not second-countable.
To do this, I'm trying to form an uncountable union $A$ of disjoint, half-open intervals of the form $[a, b)$, $a < b$. Is this possible? I think this would imply the $A$ is open but no countable union of basis elements could coincide with $A$ therefore making the real numbers with the lower limit topology not second-countable.
I think there must exist something like $A$ described above but I am having trouble visualizing it and coming up with a formula to represent it.
Maybe there is some other way to show it is not second-countable.
Best Answer
Suppose $\mathcal B$ is a base for the "lower limit" topology on $\mathbb R$, better known as the Sorgenfrey line. By the definition of a base for a topology, for any open set $U$ and any point $x\in U$ there is a basic open set $B\in\mathcal B$ such that $x\in B\subseteq U$. Hence, for any point $x\in\mathbb R$, since $[x,\infty)$ is an open set containing $x$, we can choose a set $B_x\in\mathcal B$ with $\min B_x=x$. Since the sets $B_x(x\in\mathbb R)$ are distinct, this shows that $|\mathcal B|\ge|\mathbb R|\gt\aleph_0$.