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

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.

• This space is separable however, and every separable space is ccc; that is, every collection of disjoint open sets is at most countable. We need another approach.
– user123641
Feb 6, 2015 at 5:59
• If you have an uncountable family of disjoint half open intervals, then you also have an uncountable family of disjoint open intervals (just remove the endpoint!) and there is no such thing. Feb 6, 2015 at 6:14

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

• It's a standard name by now, as Munkres uses this (and this is a popular text book). Feb 7, 2015 at 19:04
• @N.Maneesh $$\bigcup_{x\lt p\in\mathbb Q}[p,\infty)=(x,\infty)\ne[x,\infty)$$
– bof
Sep 8, 2018 at 1:59
• okay, I understood. Let me see the proof again. Thank you very much
– user464147
Sep 8, 2018 at 2:01
• Lower limit Topology seems to be more fashionable these days than Sorgenfrey Topology Feb 15, 2019 at 20:41

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 $[x,y)$.

• I'm not convinced by this proof. Wouldn't you need the same $I_{n,m}$ to work for all $x\in A$? Apr 3, 2015 at 12:12