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Topological space satisfying 2nd axiom of countability

  1. Aug 16, 2010 #1


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    Here's another problem which I'd like to check with you guys.

    So, let X be a topological space which satisfies the second axiom of countability, i.e. there exist some basis B such that its cardinal number is less or equal to [tex]\aleph_{0}[/tex]. One needs to show that such a space is Lindelöf and separable.

    To show that it's separable, let B be a countable basis for X. From every element of the basis, take one element x, and we obtain a countable set S = {x1, x2, ...}. Now we only need to show that this set S is dense in X, and this is true if and only if its intersection with every open set in X is non-empty. So, let U be some open set in X. Clearly, U can be written as a union consisting of the basis sets, and its intersection with S is definitely non-empty.

    To show that X is Lindelöf, let C be some open cover for X. The countable basis B is, by definition, an open cover, too. We only need to show that B is a refinement of C, which is obvious, because for every set U in C has a subset which belongs to B (since it can be written as a union of the basis sets).

    I hope this works, thanks in advance.
  2. jcsd
  3. Aug 17, 2010 #2
    First part looks good.

    Second part: B may be a refinement of C, but what is the countable subcover of C? In other words, why is it sufficient to find a countable refinement of C? Maybe it's because it's late, but this isn't obvious to me.
  4. Aug 17, 2010 #3


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    Hm, it seems to be a matter of definitions. In the set of lecture notes I'm working with, a space is said to be Lindelöf if every of its open covers has a countable refinement. According to this definition, X is Lindelöf.

    Then again, when speaking about compactness, in a similar manner, a topological space is defined to be compact if every of its open covers has a finite refinement. The first theorem after the definition, the one which is used frequently, states that a topological space is compact if and only if every of its open covers has a finite subcover. (Obviously, every subcover is a refinement, too.)

    So, one can conclude that a space is Lindelöf if and only if every of its open cover has a countable subcover, right? But I didn't use this fact here, it seemed more straightforward to use the definition of a Lindelöf space.
  5. Aug 17, 2010 #4
    Alright, my definition is that a space is Lindelof if every cover has a countable subcover. Your proof shows that every second-countable space satisfies your definition of Lindelof (which is presumably equivalent to mine), so it's good.
  6. Aug 17, 2010 #5


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