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Topology problem

  1. Sep 29, 2009 #1
    1. The problem statement, all variables and given/known data

    (i) Let U be a topology on Z, the integers in which every infinite subset is open. Prove that U is the discrete topology.

    (ii) Use (i) to prove that if U is a topology on an infinite sex X in which every infinite subset is open, then U is the discrete topology on X.


    2. Relevant equations

    None other than the definition of a topology.

    3. The attempt at a solution

    I have solved (i) (at least I think I have). But since X is not specified to be countable I have no idea how to apply this result to the second part. A possible idea is to consider the possibilities where X is countable and uncountable seperately and set up a bijection with Z in the countable case, but I think this is over-complicating things. I am pretty stumped here.
     
  2. jcsd
  3. Sep 29, 2009 #2

    Dick

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    Every infinite set has a countable subset, right? Just use that.
     
  4. Sep 29, 2009 #3
    Sure but does every infinite set have a countably infinite subset? This wasn't proved in my course. Even if this is true I don't really see how to use it.
    I need to show that all possible subsets of X are open given that all infinite subsets of X are open.
     
  5. Sep 29, 2009 #4

    Dick

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    Sure it's true. They should have proved it. It's basically by induction. I'm getting curious how you proved the first part. A shortcut might be to notice that if all singleton sets {x} for x in X are open, then the topology is discrete.
     
  6. Sep 29, 2009 #5
    Yeah I took the sets {...,z(k-2),z(k-1),zk} and {zk,zk+1,...} and intersected them to get {zk} which is open for arbitrary zk. Then arbitrary unions are also open hence result. I'm having trouble taking an arbitrary infinite set and singling out individual members by intersection (if this is even possible).
     
  7. Sep 29, 2009 #6

    Dick

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    Pick an x in the infinite set. There's an infinite countable subset of X that contains x. Prove it by induction. X-{x} is nonempty. Pick an x1 in it. Now X-{x,x1} is nonempty. Pick an x2 in it. X-{x,x1,x2} is nonempty. You can do this forever. The result is a countable subset of X. Now apply the first argument.
     
  8. Sep 29, 2009 #7
    I'm a little unclear on what you're proving here. You are showing that for any x in X, {x} is expressible as the intersection of two countably infinite subsets of X? That should work.
     
  9. Sep 29, 2009 #8

    Dick

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    I'm trying to show you that since every infinite set contains a countably infinite subset containing any element in X, that you can recycle your proof of the first part to prove the second part.
     
  10. Sep 30, 2009 #9
    Thanks that makes perfect sense now that I've had a break and looked again.
     
  11. Oct 9, 2009 #10
    How do you prove the first part of that question?
     
  12. Oct 10, 2009 #11
    are you saying that for the second part of the question, it is enough to say that since any infinite set contains a countably inf subset containing any element of X, it therefore must be the discrete topology of X as it is the set of all subsets? Does anything else need to be said?
     
  13. Oct 11, 2009 #12
    I am intrigued about Dick's earlier point about singleton sets. So instead of finding two infinite subsets which intersect at one point, could we just say that for every integer z, the singleton set {z} is in the topology. And that every set is a union of singleton subsets. And we know that the union of any number of sets in the topology belongs to the topology. Since the subset is arbitrary, then we must be dealing with the discrete topology?

    Or did you have something slightly different in mind Dick?
     
  14. Oct 11, 2009 #13

    Dick

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    You can't 'just say' that the singleton set {z} is in the topology. You have to prove it! Hence the point of finding two infinite sets (which are in the topology) which intersect in {x} (proving it's in the topology). The rest of your reasoning is fine.
     
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