Question concerning a statement about separability So here is my question,
Let $X$ be topological space.
If every disjoint familiy of open sets is at most countable, then $X$ is separable.
I am pretty sure that this is true but I still wanted to ask if someone knows a counter example or if someone can make sure that this is indeed true... (proof not needed)
Thanks a lot.
 A: This implication is not true.  Consider the topology on $\mathbb{R}$ where the open sets are of the form $U \setminus A$ where $U \subseteq \mathbb{R}$ is open in the usual topology, and $A \subseteq \mathbb{R}$ is countable. (I also mentioned this topology in this answer.)
It is not too hard to show that this space is c.c.c. (no uncountable families of pairwise disjoint nonempty open sets), but it is not separable (as all countable subsets are closed).

Actually, a question along these lines was very important in the development of set theory and topology.  A result of Cantor shows that $\mathbb{R}$ with the usual topology is the unique (linearly ordered) space satisfying the following properties:


*

*it has neither a greatest nor a least element;

*the order is dense;

*the order is complete;

*there is a countable dense set.


M. Souslin asked whether (4) could be replaced with 


*

*every family of pairwise disjoint open sets is countable.


and still give rise to a characterisation of $\mathbb{R}$.  It turned out that both the affirmative and negative answers to this question are consistent with the axiom of $\mathsf{ZFC}$.
