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Countable collection

  1. Feb 10, 2009 #1
    Show that the countable collection
    {(a,b) x (c,d) | a<b and c<d and a,b,c,d are rational}
    is a basis for R2.

    I was wondering... if i have to use the definition of a basis in order to solve this?
    soo... meaning.. a basis:
    Axioms:
    1. for each x [tex]\in[/tex]X, there is at least one basis element B containing x.
    2. If x belongs to the intersection of two basis elements B1 and B2, then there is a basis element B3 containing x such that B3[tex]\subset[/tex]B1[tex]\cap[/tex]B2.

    right? or am i wrong?

    Thank You,
    Jonnah Song
     
  2. jcsd
  3. Feb 10, 2009 #2

    CompuChip

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    Yes, I think that's exactly what you're supposed to do.
    In my experience, topology often makes small steps which either involve checking that the definition is satisfied, or applying the correct theorem / lemma.

    So let p = (x, y) be a point in R^2 and start by finding an element in the basis which contains p;
     
  4. Feb 10, 2009 #3

    WWGD

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    Or, if you want a more general result, prove that if the collection B is
    a basis for X , and B' is a basis for Y , then BxB' is a basis for XxY.
    It may be a good idea to try both exercises. (Note that I am not saying--
    it is actually false -- that the product topology on a product XxY is the same
    as the product of the topologies, i.e., the order topology on XxY is not necessarily
    the same as the product of the order topologies of X and Y. A specific example
    is that of the order topology on R; order intervals in R^2 are not the product
    of order intervals in R . An example when it is true, is for the discreet
    topology.)
     
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