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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.

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I do not understand your idea in just stating something and asking some body to prove or disprove.. it would be useful and moreover creates interest in people wishing to help you if you actually say why do you think so by giving at least some background for this... Hope you understand what i mean.. – Praphulla Koushik Apr 14 '14 at 13:08
@PraphullaKoushik Sorry for that. Because I think I managed to proof the statement I thought it is maybe better to hear if the statement is actually true then posting a proof for verification which for sure contains a mistake... – Thorben Apr 14 '14 at 13:12
If you post your proof you will at least know what are the mistakes in it.. Good luck anyways! – Praphulla Koushik Apr 14 '14 at 13:15
up vote 4 down vote accepted

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:

  1. it has neither a greatest nor a least element;
  2. the order is dense;
  3. the order is complete;
  4. 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 $\mathsf{ZFC}$.

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Thank you very much for the informative answer! – Thorben Apr 14 '14 at 13:18

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