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Cofinite Topology

  1. Apr 20, 2010 #1
    Hi Guys
    I was wondering if anyone knows of a good link that shows why the finite complement is a Topology?
    I been told it is the finest topology is this right?
     
  2. jcsd
  3. Apr 21, 2010 #2
    Beetle2 (Is that Paul?. Ringo.?. John.?George.?)
    " Hi Guys
    I was wondering if anyone knows of a good link that shows why the finite complement is a Topology? I been told it is the finest topology is this right? "

    I think it is a good, tho maybe a little tedious exercise in point-set topology:

    1)Let A1,A2 be in cofinite, so that X-A1, X-A2 is finite. Using DeMorgan:

    (X-A1)\/(X-A2)=X-(A1/\A2) . Can you see why X-( X-(A1/\A2)) is in (X,T), or why


    any union of sets in (X,T) is in (X,T).?


    2) (X-A1)/\(X-A2)=(X-(A1\/A2)). What happens with X-(X-(A1\/A2)).? (Can you see

    why you need finitely-many Xn's here.?
     
  4. Apr 21, 2010 #3
    Forgot to add a few things:

    I think you need to add some conditions to state whether a topology is finest

    or not (maybe with respect to some map being continuous, e.g.)

    Depending on your definition of finer topology ( unfortunately, I have seen that

    different people mean opposite things by this term), the finest topology in a

    space X is given by 2^X --the discrete topology--and the coarser one is given

    by (X, empty.)
     
  5. Apr 21, 2010 #4
    1)Let A1,A2 be in cofinite, so that X-A1, X-A2 is finite. Using DeMorgan:

    (X-A1)\/(X-A2)=X-(A1/\A2) . Can you see why X-( X-(A1/\A2)) is in (X,T)


    Is it because X-( X-(A1/\A2)) = A1/\A2 and the intercection of any closed sets in T1 is a finite set.
     
    Last edited: Apr 21, 2010
  6. Apr 21, 2010 #5
    It's straightforward, just remember De Morgan's laws for sets.
     
    Last edited: Apr 21, 2010
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