Second-countable implies separable
This article gives the statement and possibly, proof, of an implication relation between two topological space properties. That is, it states that every topological space satisfying the first topological space property (i.e., second-countable space) must also satisfy the second topological space property (i.e., separable space)
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Further information: Second-countable space
A topological space is termed second-countable if it admits a countable basis.
Further information: Separable space
A topological space is termed separable if it admits a countable dense subset.
Given: A second-countable space , with countable basis
To prove: There exists a countable dense subset of
Proof: We can assume without loss of generality that all the are nonempty, because the empty ones can be discarded. Now, for each , pick any element . Let be the set of these s. is clearly countable (because the indexing set for its elements is countable). We claim that is dense in .
To see this, let be any nonempty open subset of . Then, contains some , and hence, . But by construction, , so intersects , proving that is dense.
- Topology (2nd edition) by James R. Munkres, More info, Page 191, Theorem 30.3(b), Chapter 4, Section 30