If $X$ is a topological space such that every non-empty open subset of $X$ can be written as a countable union of disjoint open sets , then does $X$ contain a countable set (in set-theoretic sense) which is dense in $X$ (like $\mathbb Q$ in $\mathbb R$ with usual topology) ? If it is not true then does there exist any sufficient condition which ensures that a topological space contains a countable dense set ?
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4$\begingroup$ If $U$ is a set, then $U=\cup\{U\}$. So your condition in trivially satisfied. Not all topological spaces have countable dense sets. Those that do are called separable. $\endgroup$– David MitraFeb 3, 2014 at 14:10
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$\begingroup$ Maybe you wanted connected sets? $\endgroup$– Carsten SFeb 3, 2014 at 15:08
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$\begingroup$ You might want to mark this as answered if you don't have any further questions $\endgroup$– Stella BidermanMar 5, 2014 at 0:42
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As David notes, your proposed criterion is trivially satisfied for every subset of ever set. I suspect you mean being second-countable, ie having a countable base. Having a countable dense subset is known as being seperable.
Second-countable implies seperable (by virtue of the base being an example of a countae dense subset) but the reverse implication does not hold. The lower limit topology on R is seperable but not second countable. These two properties are equivalent on a metric space however. Take a countable dense subset and consider the balls of radius {1/n}. Then this forms a countable basis.
answered Feb 3, 2014 at 14:39