2
$\begingroup$

This question is basically a rerun of this one, but I am curious about its resolution and couldn't really find info online on it, and no answers were given at the other site.

The rational sequence topology on the reals (an example to be found in Steen and Seebach's "Counterexamples in Topology" book, as example 65), is is described as follows:

For every irrational number $x \in \mathbb{R}$, pick a sequence of rational numbers $(q_n(x))_{n \in \mathbb{N}}$ such that $q_n(x) \to x$ in the standard topology on $\mathbb{R}$. Define a topology on $X=\mathbb{R}$ as follows:

Every $x \in \mathbb{Q}$ has as a local base of neighbourhoods the set $\{\{x\}\}$ (so it's an isolated point), and for an irrational $x$ we have all sets of the form $U_n(x) = \{q_i(x): i \ge n\} \cup \{x\}, n \in \mathbb{N}$ as a local base of neighbourhoods.

In this topology $X$ is zero-dimensional, Hausdorff, locally compact, first countable, separable ($\mathbb{Q}$ is dense), not Lindelöf, etc.

The authors in the appendix give no reference to the definition of this space; it might well be original with them, though similar examples exist (using filters, or almost disjoint families) in the literature.

They also state problems on it in the appendix:

Problem 92: Show that in the rational sequence topology, every subset is a $G_\delta$ set.

This is indeed easy to see, I think.

Problem 93: Show that the rational sequence topology is not paracompact.

I know because of the following problem that this is the case, but

Q1: Is there a direct proof from the definition of paracompactness? An explicit cover without a locally finite refinement? EDIT now resolved by @NelsDiepeveen's comment: any cover by countable open sets will do. Any point-finite (point-countable will do too) refinement must be countable in the separable $X$, which gives us a countable subcover, and thus at most countably many reals, ouch. QED.

Problem 94: Show that for an appropriate choice of sequences, $X$ need not be $T_4$. Is this topology ever $T_4$?

This is no to the last part of the question: Jones' lemma implies that $X$ is never normal, regardless of the choice of sequences,as $X$ is separable and has a closed discrete subspace ($\mathbb{R}\setminus \mathbb{Q}$) of size $2^{\aleph_0}$. As paracompact Hausdorff implies normal, this kills paracompactness too...

Problem 95: Discuss $X$ w.r.t. whether is is ever countably paracompact for any choice of sequences.

No idea here, and this connects to the linked post on Math Stackexchange.

Here Sasha Fomin asks

Show $X$ is not metacompact, but that it is countably metacompact. Will it be countably paracompact?

I hinted that maybe the rational singletons and all $U_0(x)$ for $x$ irrational might not have a point-finite refinement, but I could not show this myself.

Q2: is this true, or do we need another cover to disprove metacompactness? EDIT This cover works, same idea as Q1...

Note that this would also answer my Q1 above. Metacompactness does not imply normality, so we have no indirect way (that I know of) to disprove it.

Q3: How to see that $X$ is countably metacompact?

EDIT
Again an enlightening comment by Niels. I worked it out in detail in my own answer to the original question.

I also had a hunch that $X$ would not be countably paracompact. But it's a meta-reason (pun intended). Dowker has shown in his famous papers on countable paracompactness that

If a space $X$ is normal, then $X$ is countably paracompact iff $X$ is countably metacompact.

And the rational sequence topology being in this counterexample book might be because the authors knew that $X$ would be an example of a space where the equivalence breaks down in a Tychonoff but non-normal space. (countably metacompact but not countably paracompact). It's weird it's not mentioned in the exercises then. Also not in the paracompactness chart on page 166, where it would have been appropriate. (Maybe they did not want to spoil their exercise?)

Q4 is this correct? Is $X$ indeed not countably paracompact? (a direct proof of this would also answer Q1 of course, in a strong way.) It might well be the case that we need to fix concrete conevrging sequences for the irrationals, in order to be able to give a concrete countable cover without a locally finite refinement (if my hunch is correct). Otherwise we'd need a better analysis of locally finite families of basic open sets?

I hope some of you can shed some lights on these questions, maybe literature references etc..

So essentially now, of the original questions just this last one remains:

Is $X$ always/sometimes (depending on the sequence choices) countably paracompact?

$\endgroup$
  • $\begingroup$ I've made all the ideas and references in @NielsJ.Diepeveen 's answer into an answer to the question that inspired it here $\endgroup$ – Henno Brandsma Jul 22 '17 at 22:09
1
$\begingroup$

Summarizing partial answers I posted earlier as comments:

Since $X$ is locally countable, $l(X) \ge |X| = \mathfrak{c}$. If $X$ were metacompact then we should have $l(X) \le d(X) = \aleph_0$. This settles questions 1 and 2.

That $X$ is countably metacompact can be shown by observing that

  • because the rationals are isolated, there is no need to worry whether they are all covered;
  • every set of irrationals is closed, so any open cover can be shrunk to one in which every irrational occurs only once;
  • if $\{U_n\}_{n<\omega}$ is a countable open cover and $\{V_n\}_{n<\omega}$ a decreasing sequence of open sets with intersection $\mathbb{R}\setminus\mathbb{Q}$, then $\{U_n \cap V_n\}_{n<\omega}$ is a sequence of open sets that covers $\mathbb{R}\setminus\mathbb{Q}$ and in which no rational $q$ occurs infinitely often, because $q\notin V_n$ eventually.

That answers question 3.

The tricky problem is whether $X$ is countably paracompact or not, or whether this depends on the choice of sequences, as the problem statement suggests it might.

Trawling through the exercises in Engelking's General Topology, I found a variation on Jones's lemma that applies to countably paracompact, rather than normal, spaces:

5.2.C.(b) [...] for every countably paracompact space $X$ and every closed discrete subspace $A$ of $X$ we have $|A| < \exp d(X).$

This implies that the rational sequence topology is never countably paracompact as easily as Jones's lemma implies that it is never normal.

By all accounts this result (actually a slightly stronger one) was first published in

Fleissner, William. "Separation properties in Moore spaces". Fundamenta Mathematicae 98.3 (1978): 279-286.

As the current, second edition of CiT was published in the same year, it is plausible that the authors simply did not know the answer at the time of writing. (The phrasing of the problem suggests they may have had a proof for a special case)

$\endgroup$
  • $\begingroup$ Nice reference find! $\endgroup$ – Henno Brandsma Jul 22 '17 at 6:53
  • $\begingroup$ BTW, It's a nice diagonalisation argument they gave, plus the good idea to switch to the $RO(X)$ base to exploit separability indirectly.. $\endgroup$ – Henno Brandsma Jul 22 '17 at 7:26
  • $\begingroup$ What goes wrong in the diagonalisation argument when we enumerate all point-finite rregular open covers instead? (so why cannot we prove that $X$ is not countably metacompact in this way?) We could still define the $Y_i$, but where does the proof break down exactly? I'm mystified... $\endgroup$ – Henno Brandsma Jul 22 '17 at 7:40
  • $\begingroup$ On more thought: I think the shrinking part of my argument breaks down. We cannot modify a point-finite refinement to a point-finite shrinking, like we can with locally finite covers. $\endgroup$ – Henno Brandsma Jul 22 '17 at 8:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.