Showing a closed subspace of a Lindelöf space is Lindelöf

  1. radou

    radou 3,108
    Homework Helper

    1. The problem statement, all variables and given/known data

    As the title says, one needs to show that if A is a closed subspace of a Lindelöf space X, then A is itself Lindelöf.

    3. The attempt at a solution

    Let U be an open covering for the subspace A. (An open covering for a set S is a collection of open sets whose union equals S, btw some define a cover for a set S as a collection such that S is contained in the union of these sets, it seems this causes disambiguity sometimes?)

    Since all elements of U are open in A, they equal the intersection of some family of open sets with A, call it U'. Now, consider the open cover for X consisting of X\A and U'. By hypothesis, this cover has a countable subcover. Dismiss X\A from it, and then the intersection of the elements left in this collection with A form a countable open cover for A.

    I hope this works.
  2. jcsd
  3. micromass

    micromass 20,069
    Staff Emeritus
    Science Advisor
    Education Advisor

    This is 100% correct!

    I'm aware that the notion of "cover" is sometimes defined in another way. But this never causes any problems. The two notions are interchangeable.
  4. radou

    radou 3,108
    Homework Helper

    Excellent! Finally a correct one. Thanks! :biggrin:

    Now back to countably dense subsets.
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