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Separable means having a countable dense set and second countable means having a countable basis.

But I cannot find any relation between dense set and basis.

bof
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user331899
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  • See this – Matias Heikkilä Oct 16 '16 at 18:37
  • You can only prove it in a metric space. In any topological space the converse is true, so in a metric space they are equivalent. – Tom Collinge Oct 16 '16 at 18:39
  • Hint: Let A be a countable dense subset of X. Take C to be the collection of all balls with rational radius around points in A. Clearly C is countable (cardinality $= |N|^2 = |N|$). Show C is a basis for the topology of X: – Tom Collinge Oct 16 '16 at 18:45
  • Can I hear someone's explanation about triangle inequality in this proof ? – user331899 Oct 16 '16 at 18:59
  • Continuing my earlier comment you use the triangle inequality to prove that C is a basis. – Tom Collinge Oct 16 '16 at 21:51
  • Are you talking about metric spaces or general topological spaces? A topological space can be separable (or even countable) without being second countable (or even first countable). But for metric spaces separable and second countable are equivalent. – bof May 21 '18 at 05:39

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It is true that all second-countable spaces are separable. The converse is not correct. For an example, take $\mathbb{R}$ in the lower limit topology, that is the one generated by intervals of the form $[a,b)$. The set $\mathbb Q$ of rational numbers is dense but it is not second-countable (show this!).

D. Brogan
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