Second-countable implies Lindelof

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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 must also satisfy the second topological space property
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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 Lindelof space.

Verbal statement

Any second-countable space is Lindelof.

Definitions used

Second-countable space

Further information: Second-countable space

A topological space is termed second-countable if it admits a countable basis.

Lindelof space

Further information: Lindelof space

A topological space is termed Lindelof if every open cover of the space has a countable subcover.

Proof

Given: A second-countable space X with countable basis \{ B_n \}

To prove: If \{ U_i \}_{i \in I} form an open cover of X, there exists a countable subcover of X among the \{ U_i \}s

Proof: For each basis element B_n, let V_n be any U_i containing B_n, if such a U_i exists, otherwise, pick nothing. This gives a (at most) countable subcollection \{ V_n \} of the collection \{ U_i \}. We want to show that this subcollection covers X.

Suppose there exists x \in X, that does not belong to any V_n. Then, since the entire collection of U_j cover X, we can find some j \in I such that x \in U_j. Further, there exists some n such that x \in B_n \subset U_j. Now, since B_n is contained in at least one of the U_is, there should exist an element V_n. But such an element would also contain x, contradicting the claim that x does not belong to any V_n.

References

Textbook references

  • Topology (2nd edition) by James R. Munkres, More info, Page 191, Theorem 30.3(a), Chapter 4, Section 30