An $L$-space is a hereditarily Lindelof regular space which is not separable.

A space is $d$-separable if it contains a dense set which is the countable union of discrete sets.

An $L$-space can't be $d$-separable, because it has no uncountable discrete subsets, however, by a small modification of his celebrated construction of a ZFC $L$-space, Justin Moore provided a ZFC example of an $L$-space with a $d$-separable square.

Moore, Justin Tatch, An $L$ space with a $d$-separable square, Topology Appl. 155, No. 4, 304-307 (2008). ZBL1146.54015.

At first glance $d$-separability looks like the strongest "separability-type" property you could hope to get in the square of an $L$-space, or is it? Consider, for example, the following "selective version" of $d$-separability:

A space is called $D$-separable if, for every sequence $\{D_n: n < \omega \}$ of dense subsets of $X$, there are discrete sets $E_n \subset D_n$, for every $n<\omega$, such that $\bigcup \{E_n: n < \omega \}$ is dense.

The above property lies between a property that the square of an $L$-space clearly can't have (a $\sigma$-disjoint $\pi$-base) and a property that the square of an $L$-space can have ($d$-separability).

QUESTION: Is there an $L$-space with a $D$-separable square?

For more information about $D$-separability see: Bella, Angelo; Matveev, Mikhail; Spadaro, Santi, Variations of selective separability II: Discrete sets and the influence of convergence and maximality, Topology Appl. 159, No. 1, 253-271 (2012). ZBL1239.54014.).


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.