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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 $B$.

Geoff Pointer
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2 Answers2

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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.

Brian M. Scott
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  • Is there a direct way you know, of proving it by demonstrating a refinement of a general open cover? I imagine that would possibly be pointless given there are probably more technically appropriate ways of doing it. Just as you're pointing out, and I am interested in pursuing topology further, but I'm currently trying to escape a sidetrack and was just curious about finding a first principles method like the one I've seen for showing $\mathbb R$ is paracompact. – Geoff Pointer Sep 26 '13 at 07:34
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    @Geoff: Direct proofs of paracompactness tend to be rather messy, so I tend not to look if an easy alternative is available. I suspect that any direct proof will essentially recapitulate at least parts of the proof that a regular Lindelöf space is paracompact or the proof that a generalized ordered space is paracompact iff it has one of a rather long list of properties known to be equivalent to paracompactness in GO-spaces. – Brian M. Scott Sep 26 '13 at 07:40
  • Looked up Lindelöf concept inspired by your answer. Using Munkres' Topology 2nd Ed. Using it as reference for homology, as cohomology is big part of my project. One text I have for that, Bott & Tu's, Differential Forms in Algebraic Topology, keeps referring to homology. My supervisor says I can avoid it but I have a pathological aversion to reading mathematics and skipping over stuff "I don't have to understand". Munkres has the Sorgenfrey Plane as $\mathbb R_\ell \times \mathbb R_\ell$ but doesn't call $\mathbb R_\ell$ the Sorgenfrey Line, just lower limit topology. Thanks for your interest. – Geoff Pointer Sep 26 '13 at 08:19
  • @Geoff: $\Bbb R$ with the lower limit topology is the Sorgenfrey line. At the risk of contributing to your tendency to indulge in cat-vacuuming, you might want to look at this post to Dan Ma’s Topology Blog; it contains most of the basic information about the Sorgenfrey line and plane. – Brian M. Scott Sep 26 '13 at 08:22
  • I knew the Sorgenfrey Line was the lower limit topology, my point is that I find it curious that Munkres doesn't call it that. "cat-vacuuming" - that's funny. Although, in my case, no cats get harmed in the process - not that I like cats that much - it's just my deadlines, grades and sleep patterns that suffer. I will read that blog you linked. – Geoff Pointer Sep 26 '13 at 08:29
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    @Geoff: Though his book is pretty good, my impression is that Munkres was/is a bit outside the mainstream of general/set-theoretic topology. Dan Ma’s Topology Blog is also pretty good; it’s definitely one of the better place to look on-line for standard results in general topology that you won’t necessarily find in undergraduate texts. – Brian M. Scott Sep 26 '13 at 08:33
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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 $\mathcal{U}$.

user642796
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