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.

  • $\begingroup$ Is there any Normal, Hausdorff, First countable counter-example ? $\endgroup$ – user Oct 15 '17 at 10:07
  • $\begingroup$ @users Yes it's in the linked paper. It needs some work :) Under CH there is a compact one, even. $\endgroup$ – Henno Brandsma Oct 15 '17 at 10:17
  • $\begingroup$ Wow , ok . What if I added Normal+Hausdorff+First countable + Paracompact ? $\endgroup$ – user Oct 15 '17 at 10:22
  • $\begingroup$ @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. $\endgroup$ – Henno Brandsma Oct 15 '17 at 10:43
  • $\begingroup$ I see ... could you please provide the construction ... ? $\endgroup$ – user Oct 15 '17 at 11:40

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.