# L-spaces without convergent sequences

An L-space is a regular hereditarily Lindelof space which is not hereditarily separable. Consistent examples of L-spaces are relatively easy to come by (for example, Suslin Lines), but the first construction of an L-space in ZFC was only given in 2006 by J.T. Moore.

Moore's space has lots of convergent sequences. Actually, it's even Frechet-Urysohn, meaning that every point in the closure of a set can be approximated by a countable convergent sequence lying inside that set (see Theorem 7.8 from Justin Tatch Moore, MR 2220104 A solution to the $L$ space problem, J. Amer. Math. Soc. 19 (2006), no. 3, 717--736 (electronic).).

However, there are consistent examples of L-spaces which lack any non-trivial convergent sequences and in a pretty extreme way. Take a Sierpinski subset of the real line (that is, an uncountable subset of $\mathbb{R}$ having countable intersection with every Lebesgue null set) and provide it with the topology inherited from the density topology (that is the topology on $\mathbb{R}$ whose open sets are the measurable sets with Lebesgue density 1 at each one of their points). Then the resulting space is regular and hereditarily Lindelof and has the property that every countable subset is closed discrete (This was first observed in H. E. White, Jr., MR 341379 Topological spaces in which Blumberg’s theorem holds, Proc. Amer. Math. Soc. 44 (1974), 454--462. and Franklin D. Tall, MR 419709 The density topology, Pacific J. Math. 62 (1976), no. 1, 275--284.).

The existence of Sierpinski set is consistent with and independent from ZFC (CH easily provides a Sierpinski set via induction on $\omega_1$ but $MA_{\omega_1}$ kills all Sierpinski sets), so it's natural to ask:

Is there in ZFC an L-space where every countable set is closed and discrete?

Is there at least a ZFC example of an L-space without any convergent sequences?

• The ideal-modification of any L-space $X$ (i.e., the new topology consists of sets which are open in $X$ modulo the ideal of countable sets) gives a hereditarily Lindelof space with closed discrete countable sets, but this space is not regular, unfortunately. – Taras Banakh Aug 30 '16 at 13:18
• Sure, and you don't even need to start with an L-space. Any X hereditarily Lindelof will do. But of course I'm looking for regular spaces. Thanks for pointing that out! – Santi Spadaro Aug 30 '16 at 23:49
• I have an honest question regarding your first question. Is it implicit that you don't want a trivial example (any countable space with the discrete topology)? Thanks. – Sergio Garcia Oct 28 '16 at 3:56
• I'm looking for hereditarily Lindelof non-hereditarily separable spaces. A countable space (with any kind of topology) is hereditarily separable. – Santi Spadaro Oct 28 '16 at 22:01