32

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.

user642796
  • 52,188
  • 2
    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. –  Feb 06 '15 at 05:59
  • 2
    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. – Mariano Suárez-Álvarez Feb 06 '15 at 06:14

2 Answers2

53

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

bof
  • 78,265
  • 7
    It's a standard name by now, as Munkres uses this (and this is a popular text book). – Henno Brandsma Feb 07 '15 at 19:04
  • 1
    @N.Maneesh $$\bigcup_{x\lt p\in\mathbb Q}[p,\infty)=(x,\infty)\ne[x,\infty)$$ – bof Sep 08 '18 at 01:59
  • okay, I understood. Let me see the proof again. Thank you very much –  Sep 08 '18 at 02:01
  • 4
    Lower limit Topology seems to be more fashionable these days than Sorgenfrey Topology – Tuo Feb 15 '19 at 20:41
  • I believe $|\mathcal{B}| =|\mathbb{R}|$ is what you were trying to write in the last line – Ryukendo Dey Apr 13 '23 at 17:24
  • @RyukendoDey It's also true that $|\mathcal B}\le|\mathcal R|$ because the collection of all open sets in the Sorgenfrey topology has cardinality $\mathfrak c$, but so what? How is that relevant to showing that the topology is not second-countable? – bof Apr 13 '23 at 22:05
  • @bof $|\mathcal{B}| = |\mathbb{R}|$ suffices for the non-second countability of lower limit topology. And I don't see how $|\mathcal{B}| > |\mathbb{R}|$ could possibly be true. You might wanna elaborate on your last sentence in your solution. Do you mean to say there is a surjective mapping from $\mathbb{R}$ to $\mathcal{B}$. If yes then I believe you need to prove it. Also please explain the meaning of the symbols $\mathcal{N}_0$ in your solution. – Ryukendo Dey Apr 14 '23 at 15:13
  • @RyukendoDey $|\mathcal B|=|\mathbb R|$ means that $|\mathcal B|\le|\mathbb R|$ AND $|\mathbb R|\le|\mathcal B|$. The former inequality is true but irrelevant. Only the latter is needed to answer the question – bof Apr 14 '23 at 21:17
  • As shown in my answer, the map $B\mapsto\min B$ is a surjection from a subcollection of $\mathcal B$ to $\mathbb R$, and the map $x\mapsto B_x$ is an injection from $\mathbb R$ to $\mathcal B$. – bof Apr 14 '23 at 21:20
  • @RyukendoDey The symbol $\mathcal N_0$ does not occur in my answer. – bof Apr 14 '23 at 21:22
  • @bof I meant $\aleph_0$ – Ryukendo Dey Apr 15 '23 at 05:38
  • @RyukendoDey $\aleph_0$ is standard notation for $|\mathbb N|$, the smallest infinite cardinal number. ($\mathbb N$ is the set of natural numbers.) I didn't think I needed to explain it because I would expect anyone interested in second-countable topologies to be familiar with it. – bof Apr 15 '23 at 05:55
  • @bof No, I didn't know this one – Ryukendo Dey Apr 15 '23 at 06:05
5

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 $[x,y)$.

Pp..
  • 5,973