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Sum of a closed set and a compact set, closed?

  1. Nov 11, 2009 #1
    1. The problem statement, all variables and given/known data
    I am trying to prove that, if X is compact and Y is closed, X+Y is closed. Both X and Y are sets of real numbers.


    2. Relevant equations



    3. The attempt at a solution
    I know that a sum of two closed sets isn't necessarily closed. So I presume the key must be the difference between the compact set and the closed set, namely the boundedness of a compact set. But I'm not sure how to exploit that fact for a proof.
     
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  3. Nov 11, 2009 #2

    Office_Shredder

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    Think about how the sum of two closed sets can fail to be closed, then explain why this can't happen if one of them is compact
     
  4. Nov 12, 2009 #3
    If X=N and Y={-n+1/n} where n is a natural number, then both are closed but the sum X+Y={1/n} is not, since it doesn't include 0 which is a limit.

    If one of them, say Y, is compact, this doesn't happen because {-n+1/n} has to be bounded by the definition of compactness, which means not all X=N is "canceled out" by the -n's in Y...

    A+B would look something like 1, 1/2, 1/3 ... 4, 5, 6, ... which is still closed.

    I feel like I only have a superficial understanding of this. I know it has to do with boundedness, but can't quite point to what it is without using a specific example... What am I missing?
     
  5. Nov 12, 2009 #4
    Hi!

    Your definition of the set X + Y doesn't correspond to my understanding. AFAIK, the standard definition is

    X + Y = {x + y : x [itex] \in[/itex] X and y [itex] \in[/itex] Y}

    With this definition, your set X + Y is still closed (since it doesn't contain the limit point 0), but it does contain points other than 1/n. Just wanted to mention this in case it helps.

    Do you know about topological groups? I ask because (R, +), with R having the usual topology, is such a group. I probably could give you some hints on your problem if you know the basic definitions and theorems in this subject. Even if not, try proving that the complement of the set X + Y is open in R.

    Let me know if any of the above is unclear, or if you want more help.

    Petek
     
  6. Nov 12, 2009 #5
    [tex]X=N[/tex]

    [tex]Y=\left\{ -n+\frac{1}{n} : n \in N \right\}[/tex]

    [tex]X+Y=\left\{\frac{1}{n} : n \in N \right\}[/tex]

    Thank you for your reply. Closed set, as far as I know, contains all of its limit points. Using that definition, my X+Y is not closed because 0 is a limit point of X+Y, but it is not contained in the set. Right?

    I haven't learned about topological groups, but I'll try working with the complement of X+Y.
     
  7. Nov 12, 2009 #6

    Office_Shredder

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    X+Y will contain for example the point 2+(-1+1)=2 The n you picked from X and the n you picked from Y don't have to be the same

    So suppose you have a sequence in X+Y that does not converge to a limit point. Think about how you can use this sequence to talk about sequences in X and Y
     
  8. Nov 12, 2009 #7
    Oh, right. I didn't catch that...
     
  9. Nov 12, 2009 #8
    Now that I think about it... X+Y is still NOT closed even if it contains points other than 1/n, isn't it? That doesn't change the fact that 0 is still a limit point of X+Y, which is not contained in the set. Finding one limit point not in the set is all it takes to make the set NOT closed, no?
     
  10. Nov 12, 2009 #9

    Dick

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    Yes, your set isn't closed. And finding one limit point not contained in the set makes it not closed. But you might be getting distracted from the original problem. If L is a limit point of X+Y then there is are sequences {xi} and {yi} such that xi+yi converges to L. If X is compact, what can you say about the sequence {xi}?
     
  11. Nov 12, 2009 #10
    If X is compact, the seq {xi} should converge to some x [tex]\in X[/tex]
     
  12. Nov 12, 2009 #11

    Dick

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    Not right. Can you rephrase that? [0,1] is compact. The sequence {0,1,0,1,0,1...} does not converge.
     
  13. Nov 12, 2009 #12
    If X is compact... there is a subsequence of {xi} that converges to a limit x in X?
     
  14. Nov 12, 2009 #13

    Dick

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    Much better. So there is a subsequence of xi+yi, which converges to L, call it xj+yj such that xj converges to M which is an element of X. Now what about yj? What must it converge to?
     
  15. Nov 12, 2009 #14
    I would say that yj has to converge to L-M... but I'm not sure how to justify that. Does it simply follow from the fact that xj+yj converges to L, and xj converges to M, so (xj+yj)-xj=yj converges to L-M?
     
  16. Nov 12, 2009 #15
    And if that is correct, it would follow that L-M is in Y since Y is a closed set. Then M+(L-M)=L is in X+Y
     
  17. Nov 12, 2009 #16

    Dick

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    Sure it does. Use epsilons and deltas if you have to. But if yj converges to L-M, then is L-M in Y? Y is closed, right?
     
  18. Nov 12, 2009 #17

    Dick

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    Yes. So X+Y is closed. Right?
     
  19. Nov 12, 2009 #18
    Right. I have kind of a silly question though... how can we assume that xi+yi converges to anything?
     
  20. Nov 12, 2009 #19

    Office_Shredder

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    We're assuming that X+Y is not closed. So there's a value
    L such that xi+yi converges to L (xi+yi is some sequence of points in X+Y). We're assuming we pick a sequence that converges
     
    Last edited: Nov 13, 2009
  21. Nov 13, 2009 #20

    Dick

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    Go back to post 9. We are taking L to be a limit point of X+Y because we want to show X+Y contains all of it's limit points. What's the definition of 'limit point'?
     
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