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given a topological space $(X, \tau)$, if there are uncountably infinite disjoint open subsets, then one can conclude that $(X, \tau)$ is not separable (since any dense subset of $X$ must intersect with every open subset of $X$), is the converse true?(i.e. there must be uncountably infinite disjoint open subsets in an non-separable space.)

what happens if the topological space is replaced with a metric space $(X, d)$(consider the topology induced by $d$)

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The converse is not true: https://topology.pi-base.org/theorems/T000021/converse

However it becomes true when you add the metrizable assumption: https://topology.pi-base.org/spaces?q=Ccc%2B~Separable%2BMetrizable

Finally you may find this interesting: https://en.wikipedia.org/wiki/Suslin%27s_problem#Formulation In short, whether every complete, dense, ccc linear order is separable is independent of the usual axioms of set theory.

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  • $\begingroup$ Nice answer! And thank you for mentioning the pi-base. That's a wonderful website! $\endgroup$
    – Jason Chen
    Commented Dec 29, 2023 at 9:58

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