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A problem in real analysis

  1. Aug 22, 2011 #1
    1. The problem statement, all variables and given/known data

    Suppose A is a compact set and B is a closed subset of Rk. then A+B is closed in Rk. show that A+B for two closed sets is not necessarily closed by a counter-example.

    well, since A is a compact set and there's a theorem in Rudin's mathematical analysis chapter 2 stating that compact sets of metric spaces are closed, we can conclude that A is closed. that means every limit point of A is in A. B is closed by assumption, and that means every limit point of B is in B. now here is where I get confused. is A in Rk? I think it should be because if not then how A+B is defined? I want to define A+B as the set {a+b: A,B are in Rk} but the problem has ambiguity and doesn't precisely say what A is. Do we need to know what A is? or we can solve it without knowing what A is? It's easy to solve it if A is in Rk, but I don't know what I should do if A is not in Rk.
     
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  3. Aug 22, 2011 #2

    Dick

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    A is in R^k. As you said, if it's not then A+B doesn't make much sense.
     
  4. Aug 22, 2011 #3

    hunt_mat

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    Dick, perhaps it the set defined as [itex]A+B:=\{a+b|a\in A,b\in B\}[/itex]
     
  5. Aug 22, 2011 #4
    Yea. probably. I guess that's the only way that it can make sense.

    Yea but how do you define a+b? imagine that a is a scalar quantity in R and b is a vector in R3. how do you define a+b? the question is not how A+B is defined, it's how a+b is defined.

    Assuming that A is a subset of Rk and Assuming lima=p and limb=q, can I say that A+B is closed because for every a+b we have lim(a+b) = lima + limb=p+q which is an element of A+B? (because p is in A and q is in B since A,B are closed sets)
     
  6. Aug 22, 2011 #5

    micromass

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    No. You must prove that if [itex](a_n+b_n)[/itex] converges to z, then z in A+B. But you don't know if [itex](a_n)_n[/itex] or if [itex](b_n)_n[/itex] converges...
    So you can't assume lim(a)=p and lim(b)=q...

    HOWEVER!! What does compactness of A tell you about the sequence [itex](a_n)_n[/itex]?? What does it have?
     
  7. Aug 22, 2011 #6
    There is a theorem in Rudin that says if p is a limit point of E, then there is a sequence in E that converges to p. now if a+b is a point of A+B, then by definition of A+B, a is a point of A that means there's a sequence in A that converges to a, also b is a point of B that means there's a sequence in B that converges to b. now can't I say that a+b is also a limit point because of the argument I said earlier?

    compactness of A tells me that for each of A's open covers there exists a finite sub-cover.
     
  8. Aug 22, 2011 #7

    micromass

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    Hmm, actually I'm referring to theorem 3.6. But did you see that already?
     
  9. Aug 22, 2011 #8
    Now I see. and to prove the converse, A and B must be complete but we don't know if they are complete or not. am I right?

    What does compactness say about sequences?? Apply theorem 2.37 to [itex]\{a_n~\vert~n\}[/itex][/QUOTE]

    theorem 2.37 says if E is an infinite subset of a compact set K, then E has a limit point in K. since {a(n)|n} is a subset of A and A is compact and also {a(n)|n} is an infinite set then the theorem assures us that a(n) has a limit point in A. so? I'm not quite understanding your next step.
     
  10. Aug 22, 2011 #9

    micromass

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    Right. Well, we do know that A is complete since it is compact, but we know nothing of B.

    I meant theorem 3.6 (which follows from 2.37).

    Let me reformulate.

    We have to show that if z is a limit point of A+B, that then z is in A+B, ok?
    Well, if z is a limit point, then there exists a sequence [itex]a_n+b_n\rightarrow z[/itex].
    You have already shown that if [itex](a_n)_n[/itex] and [itex](b_n)_n[/itex] converge, that z is indeed in A+B.

    The problem is that [itex](a_n)_n[/itex] doesn't need to converge. That is where compactness comes in. Compactness tells us that there exists a subsequence [itex](a_{k_n})_n[/itex] which DOES converge!! Can you know
    1) show that also [itex](b_{k_n})_n[/itex] converges??
    2) modify your argument above to conclude that z is in A+B?
     
  11. Aug 22, 2011 #10

    hunt_mat

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    I am assuming both A and B are subsets in R^k, so a+b would be simple vecor addition.
     
  12. Aug 22, 2011 #11
    You mean theorem 3.6 part a?

    Yea. and we have to prove this regardless of assuming that a(n) and b(n) are convergent sequences.

    Actually I looked at the theorem you referred me to. It states that "if {pn} is a sequence in a compact metric space X, then some subsequence of {pn} converges to a point of X". it doesn't precisely say if that point is a particular point of X or can be any point of X.maybe every sequence in A is convergent but converges to only one point of A? Isn't that possible? then there are points of A that there is no sequence that converges to them and my argument fails again.

    I don't know how to do that. the problem tells us only that it's closed.

    This is the result of my new thoughts about the problem. assume that a+b is a limit point of A+B. that means every neighborhood of (a+b) contains an infinitely many points of A+B. if it contains an infinitely many points of A+B then it also contains an infinitely many points of a & b by definition of A+B (Is it true? I guess It's true because if it doesn't, let's say if it contains only a finitely many points of A, then a+b's won't be infinitely many). therefore a+b is also a limit point of A and B and that means I can find sequences in A and B that converge to the points a and b in A and B. that means I can again use my previous argument and deduce that a+b is in A+B. is this a valid argument?

    Yea but my first question was whether A is a subset of R^k or not. Thanks for the help anyways.
     
  13. Aug 22, 2011 #12

    micromass

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    Yes.

    I don't see what you're worrying about here. You know that [itex](a_n)_n[/itex] has a convergent subsequence [itex](a_{k_n})_n[/itex]. And you still know that

    [tex]a_{k_n}+b_{k_n}\rightarrow z[/tex]

    (with z being our limit point). So can you show [itex]b_{k_n}[/itex] to be convergent as difference of [itex](a_{k_n}+b_{k_n})_n[/itex] and [itex](a_{k_n})_n[/itex]??


    It already breaks down here. You do not know that a limit point of A+B is of the form a+b. Showing that it is of the form a+b is exactly what you need to show!!
     
  14. Aug 22, 2011 #13
    Ah right. then [itex]b_{k_n}[/itex] = [itex](a_{k_n}+b_{k_n})_n[/itex] - [itex](a_{k_n})_n[/itex] is convergent too. Now we're done. right?
    but we have nowhere used the fact that A,B are subsets of R^k. Does that mean we can generalize our problem to other metric spaces as well provided we can define A+B?


    Okay right. Thanks for pointing that out.
     
  15. Aug 22, 2011 #14

    micromass

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    Yes, this is also true in the more general case! In fact, I think that this is true for arbitrary topological groups, but that requires a quite different proof, as we cannot work with sequences there.
     
  16. Aug 22, 2011 #15
    I'm reading the whole topic now to come up with a full argument. first I've shown that if a(n) and b(n) converge to p and q, then p+q is a limit point of A+B which is of the form a+b and is indeed in A+B.
    later, we've tried to show that if z is a limit point, then we can find two convergent sequences in A,B that their sum converges to z. but we don't know what a(n) and b(n) converge to. we only know that their sum converges to z, but we don't know what each of these sequences converge to alone. right?
     
  17. Aug 22, 2011 #16

    micromass

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    That would be a good summary, yes!

    Furthermore, we don't even know if a(n) and b(n) converge!! They might even diverge
     
  18. Aug 22, 2011 #17
    That's exactly what I was thinking about. How have we taken care of that case? I mean we bothered ourselves to show that a(n) converges and then concluded that b(n) converges too. but why did we do that if they can both be divergent and yet their sum converges to z?
     
  19. Aug 22, 2011 #18

    micromass

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    No, we never showed that a(n) converges. We showed that a subsequence of a(n) converges!! And it is sufficient to do everything with this subsequence!
     
  20. Aug 22, 2011 #19
    Yes, I know what you mean. but when did we show that a(n)+b(n)->z even if a(n) and b(n) diverge?
     
  21. Aug 22, 2011 #20

    micromass

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    We assumed a(n)+b(n)-> z.

    We took z to be a limit point, thus there existed a sequence in A+B that converges to z. This sequence can be taken as a(n)+b(n)...
     
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