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Homework Help: An analysis question

  1. Aug 16, 2011 #1
    1. The problem statement, all variables and given/known data

    consider the set A={1/n+1/m: n,m are natural numbers}. find A'

    3. The attempt at a solution

    I don't know what the problem is asking me. I don't know what A' represents here. I assume that A' stands for the set of the limit points of A because this is an Analysis problem and Rudin's Analysis use that notation for the set of all limit points. but I'm not 100% sure. Does anyone know what A' represents here?
    If A' stands for the set of all limit points, then what should I do to find all the limit points of A? I know that 0 is a limit point of the set, because any open ball (with arbitrary radius) centered at 0 contains an infinite number of points in the neighborhood and I can show that in a fairly easy way. but I don't know whether there are other limit points or not. I guess 0 is the only limit point of the set if the metric space that we're working with is Real numbers. but how can I prove that? I guess I need to show that for any other real number, I can find delta such that the neighborhood of that number within the radius delta contains only a finite number of points in A and therefore that real number can not be a limit point of A. am I right?
     
  2. jcsd
  3. Aug 16, 2011 #2
    May it be the complementary set of A ?
     
  4. Aug 16, 2011 #3
    I doubt that because in the same problem set it uses Ac for A compliment.
     
  5. Aug 16, 2011 #4

    HallsofIvy

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    Yes, A' is the set of limit points of set A (. If A has no "isolated points" then A' is also the closure of A. Note that 1/n+ 1/n= (n+ m)/mn. Now, if a and b are any two integers, under what conditions on a and b must there exist integers, m, n such that n+ m= a and mn= b?
     
  6. Aug 16, 2011 #5
    well,I have no knowledge of number theory and I know only some few things from naive number theory backing high school but we can solve n^2 - an + b and find if it has any integer roots. by quadratic formula, we have: n = (a[itex]\pm[/itex][itex]\sqrt{a^2 - 4b}[/itex])/2. if we want n to be an integer then a^2-4b must be a perfect square. if a^2-4b=k^2 and a[itex]\pm[/itex]k is even then n exists. if n exists, then m exists if and only if n|b. am I right?

    but I don't quite understand why we need to know that. how is that relevant to finding the limit points of A?
     
  7. Aug 17, 2011 #6
    Any ideas?
     
  8. Aug 17, 2011 #7
    Can you show that things of the form 1/n are also in A'??

    In general, you must find the limits of all sequences in A. So can you show that all sequences in A either converge to 1/n or to 0?
     
  9. Aug 17, 2011 #8
    Do you mean that I have to fix one of the two fractions like 1/m and show that the other one forms a sequence that I can converge to any term of it I want? I think I'm lost.
     
  10. Aug 17, 2011 #9
    Well, perhaps you can begin by showing that it suffices to look at sequences

    [tex]\frac{1}{n_0}+\frac{1}{m_0}, \frac{1}{n_1}+\frac{1}{m_1}, \frac{1}{n_2}+\frac{1}{m_2}, \frac{1}{n_3}+\frac{1}{m_3},...[/tex]

    such that [itex]n_0\geq n_1\geq n_2\geq n_3\geq ...[/itex] and [itex]m_0\geq m_1\geq m_2\geq m_3\geq ...[/itex]?

    By the way, did you already show that 1/n is a limit point for each n?
     
  11. Aug 17, 2011 #10
    I'm not quite understanding what you're explaining.

    I guess you mean that 1/n is a limit point for each n because I can always choose m sufficiently large in such a way that any neighborhood centered at 1/n contains an infinite number of points of A?
    for example if I want to show that 1/5 is a limit point of the set, I can see that if I choose the radius of the neighborhood to be 0.5(just as an example) then by choosing m>=2 then N(1/n,0.5) contains an infinite number of points of A. am I right?
     
  12. Aug 17, 2011 #11
    Yes, you are right, but try to do it with sequences. Can you find a sequence in A that converges to 1/n?
     
  13. Aug 17, 2011 #12
    a(m)=1/m+1/n. (a: N-> A) you mean a sequence like this? does this work?
     
  14. Aug 17, 2011 #13
    Yes, this works.
    Now try to show (with sequences) that no other point is a limit point.
     
  15. Aug 17, 2011 #14
    then I guess that zero is not a limit point. am I right?
    Now I intuitively know what we're doing, but I'm still a little bit confused. Do you mean to show that x is a limit point of A I should show that there exists a sequence a:N -> A that maps every natural number to a point of A?
    How can I show that there is no other limit point of A except the ones that are of the form 1/n? It seems somehow obvious now, but I'm having problem with giving a formal argument.
     
  16. Aug 17, 2011 #15
    Nono, 0 IS a limit point of course. 0 and 1/n are all the limit points. Sorry if I confused you.

    Now I intuitively know what we're doing, but I'm still a little bit confused. Do you mean to show that x is a limit point of A I should show that there exists a sequence a:N -> A that maps every natural number to a point of A?
    How can I show that there is no other limit point of A except the ones that are of the form 1/n? It seems somehow obvious now, but I'm having problem with giving a formal argument.[/QUOTE]

    You must show that every convergent sequence of terms in A converges to either 1/n or 0. So take a convergent sequence in A, this has the form

    [tex]\frac{1}{n_0}+\frac{1}{m_0},\frac{1}{n_1}+\frac{1}{m_1}, \frac{1}{n_2}+\frac{1}{m_2},\frac{1}{n_3}+\frac{1}{m_3},...[/tex]

    and see if you can prove that this converges to either something of the form 0 or 1/n.
     
  17. Aug 17, 2011 #16
    You must show that every convergent sequence of terms in A converges to either 1/n or 0. So take a convergent sequence in A, this has the form

    [tex]\frac{1}{n_0}+\frac{1}{m_0},\frac{1}{n_1}+\frac{1}{m_1}, \frac{1}{n_2}+\frac{1}{m_2},\frac{1}{n_3}+\frac{1}{m_3},...[/tex]

    and see if you can prove that this converges to either something of the form 0 or 1/n.[/QUOTE]

    I guess I'm still confused for 3 reasons:
    1- Rudin defines that p is a limit point of the set E if every neighborhood of p contains a point q[itex]\neq[/itex]p such that q[itex]\in[/itex]E. therefore a limit point of a set may not be an element of it. am I right?
    2- I don't understand how I should deal with that sequence. we got two variables in that sequence and I don't know how to handle it. I also have not taken multi-variable calculus yet but I guess I should keep one constant and vary the other one. dealing with a 2 variable sequence is the part that I get confused.
    3- I don't understand how I'm supposed to show that 1/n and 0 are the only limit points of A using sequences. and also I've not reached chapter 3 of Rudin's analysis yet to know if Rudin has provided any theorems or methods about sequences that can be used for finding limit points of a set.
     
  18. Aug 17, 2011 #17
    I guess I'm still confused for 3 reasons:
    1- Rudin defines that p is a limit point of the set E if every neighborhood of p contains a point q[itex]\neq[/itex]p such that q[itex]\in[/itex]E. therefore a limit point of a set may not be an element of it. am I right?
    [/QUOTE]

    No, that is not correct. A limit point can be an element of the set. A point p is a limit point of E if every neighborhood around p contains a point in E different from p. This does not exclude p from being an element of E.
    For example, the limit points of the open interval ]0,1[ is [0,1]. Try to verify that.

    An analogous statement is that p is a limit point of E, if there exists a sequence in E\{p} that converges to p.

    Well, forget about the two variable sequence for a while then. Suppose that we have a sequence [itex](x_n)_n[/itex] in A that converges to a point. Our job is to show that the point is 1/n or 0. Now, showing this immediately can be difficult, so let's first simplify our task. Can you show that it sufficed to take [itex](x_n)_n[/itex] a decreasing sequence?? I.e. can you show that a strictly increasing sequence in A cannot converge?

    (hint: try to show first that an strictly increasing sequence in {1/n | n>0} cannot converge)
     
  19. Aug 17, 2011 #18
    I know that it can be an element of the set as well. but even if it wasn't, there would be no problems. right? because I used to think that it must be an element of E and that made me confused.

    when we say a sequence{xn} is in A that means every xn is in A. am I right? here is where I get confused again. if we say that 0 is a limit point of A, then there must be a sequence from N -> A that converges to 0. but 0 is obviously not in A. so how can any sequence in A converges to an element that doesn't exist in the set?

    I don't understand what you're saying at all :D This is the first time that I feel I don't understand a word of something about math :D I'm confused xD
     
  20. Aug 17, 2011 #19
    Well, the sequence 1/n converges to 0, right?? And 0 is not in A?

    Welcome to Rudin.
     
  21. Aug 17, 2011 #20
    Yea, but How is that possible? Can we have a sequence in real numbers that converges to a complex number?

    hahaha. This sequence thing was phreaking confusing because I kept getting confused what we were talking about lol. Would you solve this particular problem for me that I read it and see if I can understand it when I have the proof?
     
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