X is separable if there is a countable dense subset in X. X is local separable if every point has separable neighborhood.
The following example is standard: take an uncountable
set X with discrete topology. Then each point is its own neighbourhood,
so the space is locally seperable at every point. But it is not separable.
I'm not sure about that theorem attributed to Sierpinski: If X is the disjoint union of two separable open sets U and V, doesn't that make X separable? BTW, I agree with Ruzinazar; indeed any locally separable Lindelof space is separable.
There must be something wrong. The disjoint sum of finitely (even countable) many separable spaces is separable.
Or there may be the disjoint sum of separable open sets (NOT TWO ! ).
It is true that a locally separable Lindelof space is separable. However, a space is separable if and only if it is locally separable and has the countable chain condition.
The countable chain condition (ccc for short) means that every family of, disjoint nonempty open sets, is countable .
It is true that a locally separable Lindelof space is separable. However, a space is separable if and only if it is locally separable and has the countable chain condition.
The countable chain condition (ccc for short) means that every family of, disjoint nonempty open sets, is countable .
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