# Which non-separable topological spaces has an uncountable family of pairwise disjoint open sets?

If $(X,d)$ is a non-separable metric space then I can show that there exist an uncountable set $S \subseteq X$ and $r>0$ such that $d(x ,y ) >r , \forall x, y \in S , x \ne y$ ; hence there exist an uncountable collection of pairwise disjoint open sets of $X$ . Now this is not necessarily true for general topological spaces : $\mathbb R$ with co-countable topology is not separable but any two non-empty open sets in this topology intersect .

My question is : Under some separation and/or countablity axiom , can we say that non-separability of topological space implies the existence of an uncountable family of pairwise disjoint open sets ?

There are 3 notions from general topology that are involved here:

1. $X$ is called separable if it has a countable dense set.
2. $X$ is called ccc if every family of non-empty pairwise disjoint open sets is at most countable.
3. $X$ has countable spread iff every discrete subspace is at most countable.

You're basically asking when a ccc space is separable (which is equivalent logically to non-separable implies non-ccc).

In metric spaces, these 3 properties are equivalent (to being second countable, or Lindelöf etc.). In any space separable implies ccc, but there are quite a few counterexamples (ccc non-separable spaces) in general spaces, like large products $[0,1]^I$ where $|I| > |\mathbb{R}|$, e.g., or the co-countable topology (as you noted), or the one-point compactification of an uncountable discrete space. The first and third are compact Hausdorff, but not first countable, but there are also consistent counterexamples that are close to $\mathbb{R}$ (ordered, first countable dense order etc.; Look up Suslin lines), so no obvious extra assumptions are enough to go back from ccc to separable I think. (well second countable implies both but that's trivial).

Murray Bell constructed in 1979 the first ZFC example of a normal first countable ccc non-separable space. MA + non-CH implies that all perfecty normal ccc spaces are separable, so in that case we can reverse the implication if we are willing to assume an extra axiom. ccc and non-separable spaces were a hot topic in topology for some time.

• Is there any Normal, Hausdorff, First countable counter-example ? – user Oct 15 '17 at 10:07
• @users Yes it's in the linked paper. It needs some work :) Under CH there is a compact one, even. – Henno Brandsma Oct 15 '17 at 10:17
• Wow , ok . What if I added Normal+Hausdorff+First countable + Paracompact ? – user Oct 15 '17 at 10:22
• @users I can make (under axioms somewhat weaker than CH) an example of a compact (so certainly paracompact) Hausdorff first countable ccc (even hereditarily Lindelöf) non-separable space. – Henno Brandsma Oct 15 '17 at 10:43
• I see ... could you please provide the construction ... ? – user Oct 15 '17 at 11:40