4
$\begingroup$

Recall that a space is:

  • "Lindelof", if every open cover has a countable subcover.
  • "Linearly Lindelof", if every open cover which is linearly ordered by $\subseteq$ has a countable subcover.
  • "weakly Lindelof" if every open cover has a countable subcollection whose union is dense in the space.

Of course every Lindelof space is both weakly Lindelof and Linearly Lindelof. It's easy to see that every space with the countable chain condition (i.e., "there are no uncountable families of pairwise disjoint non-empty open sets") is weakly Lindelof and that every space with a Lindelof dense subset is weakly Lindelof. Using this observation one can easily find a weakly Lindelof non-Lindelof space: any separable non-Lindelof space, like the tangent disk plane, or the Cantor tree, will do. These examples are also non-linearly Lindelof, since they contain an uncountable closed discrete subspace.

Finding a Linearly Lindelof non-Lindelof space is harder, and one of the most outstanding problems in set-theoretic topology is whether there exists a normal Linearly Lindelof space which is not Lindelof. A positive answer to this problem would lead to a new construction of a Dowker space, that is a normal space whose product with the unit interval is not normal.

Is there a Linearly Lindelof Tychonoff space which is not weakly Lindelof?

All known examples of Linearly Lindelof non-Lindelof spaces seem to be weakly Lindelof. Here are the two most popular ones:

  • Mishchenko space is $X=\bigcup_{n \in \mathbb{N}} (\prod_{k \leq n} (\omega_k+1) \times \prod_{k>n} \omega_k)$ with the topology inherited from $\prod_{n \in \mathbb{N}} (\omega_n+1)$. This is weakly Lindelof because it has a $\sigma$-compact dense subspace.
  • (Buzyakova and Gruenhage) Denote with $2^{\aleph_\omega}$ the product of $\aleph_\omega$ many copies of the 2-point discrete space. Let $X=\{x \in 2^{\aleph_\omega}: |x^{-1}(1)| < \aleph_\omega \}$ with the topology inherited from the usual product topology on $2^{\aleph_\omega}$. This example is weakly Lindelof for two reasons: first of all, it also has a $\sigma$-compact dense subspace and then it has the countable chain condition.

Most other examples are obtained by tweaking one of those two to get some additional special property, like first-countability (Pavlov), hereditary realcompactness (Arhangel'skii and Buzyakova), etc...

$\endgroup$
  • $\begingroup$ What is a linearly ordered open cover? Is it just an open cover together with a linear ordering on it? Axiom of choice implies every set can be linearly ordered, so I am probably missing something. $\endgroup$ – Wojowu May 3 at 11:26
  • 2
    $\begingroup$ Linearly ordered by containment. $\endgroup$ – Santi Spadaro May 3 at 11:26
  • $\begingroup$ That makes more sense, thanks. $\endgroup$ – Wojowu May 3 at 11:27

Your Answer

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

Browse other questions tagged or ask your own question.