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Lindelof Space.

  1. Apr 22, 2008 #1


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    I need to show that if X is Lindelof and Y is compact then XxY is Lindelof.
    Now let ,[tex]XxY=U(A_\alpha x B_\beta)[/tex] for arbitrary alpha and beta.
    from naive set theory we know that XxY is a subset of [tex]UA_\alpha x UB_\beta[/tex]
    because both UA_alpha and UB_beta are open in X and in Y respectively we get that XxY equals the last term.
    Now from X being Lindelof we get from the above that X=UA_alpha, and from Lindelof we know there's a countable covering of X now also Y equals UB_beta and from compactness we know it has a finite covering of B_i's.
    so [tex]XxY=UA_j x UB_i[/tex]
    where UA_j is the countable covering of X and UB_i is a finite covering of Y, but this same set also equals: U[(A_j x B_1)U(A_j x B_2)U(A_j x B_n)] where j runs from one to infinity, this is a countable covering of XxY, or so I think cause A_j xB_i for every j>=1 and some i natural in [1,n], is open in XxY.
    What do you think, looks sound and reasonable or bullocks?
    the small x's in the texs are for cartesian product and capital U'S are for unions, sorry don't have time to relearn Latex, I have 3 exams in four days, wish me luck.
    Last edited: Apr 22, 2008
  2. jcsd
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