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...