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)
View all topological space property implications | View all topological space property non-implications
Get more facts about second-countable space|Get more facts about separable space
Contents
Statement
Property-theoretic statement
The property of topological spaces of being a second-countable space implies, or is stronger than, the property of being a separable space.
Verbal statement
Every second-countable space is a separable space.
Definitions used
Second-countable space
Further information: Second-countable space
A topological space is termed second-countable if it admits a countable basis.
Separable space
Further information: Separable space
A topological space is termed separable if it admits a countable dense subset.
Proof
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.
References
Textbook references
- Topology (2nd edition) by James R. Munkres, ^{More info}, Page 191, Theorem 30.3(b), Chapter 4, Section 30