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Showing a closed subspace of a Lindelöf space is Lindelöf

  1. Nov 17, 2010 #1

    radou

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    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. Nov 17, 2010 #2

    micromass

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    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. Nov 17, 2010 #3

    radou

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

    Now back to countably dense subsets.
     
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