$\mathbb{R}$ with the lower limit topology is not second-countable

Question

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.

Answer 1

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 $|\mathcal B|\ge|\mathbb R|\gt\aleph_0$.

Answer 2

Let $A_n$ be such a base. For each $A_n\subset A_m$ pick, if possible, an interval $I_{n,m}:=[a_{n,m},b_{n,m})$ such that $A_n\subset [a,b)\subset A_m$.

The collection $I_{n,m}$ should be a base. In fact, if $A$ is open, take and arbitrary point $x\in A$. Then there is (because the $A_n$ for a base) an $A_n\ni x$ contained in $A$. There is (because the $[a,b)$ form a base) a small $[a,b)\ni x$ contained in $A_n$ and there is (because $A_n$'s are a base) an $A_m\ni x$ contained in $[a,b)$. Therefore there is an $I_{n,m}$ (because for this particular $n,m$ is is possible such $I_{n,m}$. Notice that $[a,b)$ could be a candidate). This $I_{n,m}$ incidentally contains $x$ and is inside $A$. Therefore $A$ is equal to the union of all the $I_{n,m}$ inside it.

Above we didn't really used anything about the particular bases. In general: Given two bases you can construct a subset of one, that is still a basis, and has cardinality not larger than the other base.

Therefore there is a countable base $I_{n,m}=[a_{n,m},b_{n,m})$. Since $\mathbb{R}$ is uncountable, there are two points $x,y$ such that they are not boundaries of any $I_{n,m}$.

But $[x,y)$ can't be formed by a union of $I_{n,m}$, the point $x$ is never covered by the $I_{n,m}$ lying inside the interval $[x,y)$.