Showing the Sorgenfrey Line is Paracompact

Question

The Sorgenfrey Line is $\mathbb R_/ = (\mathbb R, \tau_s)$ where $\tau_s$ is the topology on $\mathbb R$ with base $\{[a, b)\ |\ a, b \in \mathbb R\}$. I know how to show $\mathbb R_/$ is not locally compact. It turns out that the only compact sets in $\mathbb R_/$ are at best countable.

What I want to prove is that it is paracompact. You can't use the usual proof that works on $\mathbb R$ because that uses a refinement of any open cover of $\mathbb R$ that is constructed using the fact that closed balls are compact in $\mathbb R$, which can't be used here.

I'm sure I'll figure this out eventually but any feedback would be good. This is not a homework question, it's a sidetrack from my Honours project, where I have to use paracompactness as a prerequisite in some of the proofs I'm studying. I got sucked into studying the Sorgenfrey line, but I need to get back to my project, I'm too easily distracted.

Glossary:

Locally Compact: $X$ is locally compact if for every $x \in X$ there exists a compact set $K \subset X$ that itself contains an open neighbourhood of $x$.

Paracompact: $X$ is paracompact if every open cover $A$ of $X$ has a refinement $B$ so that every $x \in X$ has an open neighbourhood that intersects with finitely many members of $B$.

Answer 1

HINT: Prove that the Sorgenfrey line is Lindelöf, and use the fact that a regular Lindelöf space is paracompact. To prove that it’s Lindelöf, start with a basic open cover $\mathscr{U}$ (i.e., a cover by sets of the form $[a,b)$). Show that $\{(a,b):[a,b)\in\mathscr{U}\}$ covers all but a countable subset of $\Bbb R$, and use the fact that $\Bbb R$ with the usual topology, being second countable, is hereditarily Lindelöf.

Answer 2

Late answer, but I thought I'd add a "getting your hands dirty" approach. Below I will denote by $\mathbb{S}$ the Sorgenfrey line, by $\mathbb{R}$ the real line (with the usual topology), and by $\mathsf{R}$ simply the set of real numbers. If I give a subset of $\mathsf{R}$ the superscript either $^{\langle \mathbb{S} \rangle}$ or $^{\langle \mathbb{R} \rangle}$ I intend to consider it as a subset (or subspace) of the Sorgenfrey line or the real line, respectively. (So, for example, "$A^{\langle \mathbb{S} \rangle}$ is open" means that $A$ is an open subset of $\mathbb{S}$.)

1. Let $\mathcal{U}$ be an arbitrary open cover of $\mathbb{S}$.

2. Consider $W := \bigcup_{U \in \mathcal{U}} \operatorname{Int}_{\mathbb{R}} ( U )$ (where the interior is taken with respect to $\mathbb{R}$). Clearly, $W^{\langle \mathbb{R} \rangle}$ is open, and so it is a countable union of pairwise disjoint open intervals. Without any essential loss of generality, we may assume that each of these open intervals is bounded, i.e., $W = \bigcup_{n \in \mathbb{N}} (a_n,b_n)$, and $(a_m,b_m) \cap (a_n,b_n) = \varnothing$ whenever $m \neq n$.

3. We can show that $\mathsf{R} \setminus W = \{ a_n : n \in \mathbb{N} \}$.

4. Given any $n \in \mathbb{N}$:

   • There is a $U_n \in \mathcal{U}$ and a $d_n > a_n$ such that $[a_n,d_n) \subseteq U_n$. Note that it must be that $d_n \leq b_n$. (If $d_n = b_n$ we can set $\mathcal{W}_n := \varnothing$ and skip the next two points for this $n$; we thus assume that $d_n < b_n$ in the next two points.)

   • Consider $\mathcal{W}_n^* := \{ \operatorname{Int}_{\mathbb{R}} ( U ) \cap (a_n,b_n) : U \in \mathcal{U}, U \cap (a_n,b_n) \neq \varnothing \}$. It follows that $\mathcal{W}_n^*$ is an open cover of $(a_n,b_n)^{\langle \mathbb{R} \rangle}$, and since $(a_n,b_n)^{\langle \mathbb{R} \rangle}$ is paracompact, there is a locally finite open refinement $\mathcal{W}_n^\prime$ of $\mathcal{W}_n^*$.

   • Now let $\mathcal{W}_n := \{ W \cap [ d_n , b_n ) : W \in \mathcal{W}_n^\prime \}$. It follows that $\mathcal{W}_n$ is a locally finite family of open subsets of $\mathbb{S}$ which covers $[d_n,b_n)$.

5. It can then be shown that $\mathcal{V} := \{ [a_n,d_n) : n \in \mathbb{N} \} \cup \bigcup_{n \in \mathbb{N}} \mathcal{W}_n$ is a locally finite open refinement of $\mathcal{U}$.