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Topology & Compactness in R^n

  1. Sep 30, 2007 #1
    1) Prove rigorously that S={(x,y) | 1< x^2 + y^2 <4} in R^n is open using the following definition of an open set:
    A set S C R^n is "open" if for all x E S, there exists some r>0 s.t. all y E R^n satisfying |y-x|<r also belongs to S.


    [My attempt:
    Let x E S, r1 = 2 - |x|, r2= |x| - 1, r = min {r1,r2}
    Let y E R^n s.t. |y-x|<r
    =>|y-x|<r1 and |y-x|<r2

    |y|<=|y-x|+|x| (triangle inequality) [<= means less than or equal to]
    <r1+|x|=2-|x|+|x|=2
    So |y|<2

    Now if I can prove that |y|>1, then I am done (y belongs to S). However, I tried many different ways, but still unable to prove that |y|>1, what should I do? ]


    2) Let Li denote the line segment in R^2 from the origin (0,0) to the point (1/i, sqrt(1/i) ) on the curve f(x)=sqrt x. Prove that the infinite union S=U(i=1 to infinity) Li is compact using the Bolzano-Weierstrass Theorem.

    Bolzano-Weierstrass Theorem: Let S C R^n. Then S is compact (bounded and closed) iff every sequence of points in S has a convergent subsequence whose limit lies in S.


    Any help, explanation, or hints? I am feeling totally blank on this question...


    My textbook basically has no examples, so I don't know how to use the theorems and I am really frustrated when doing the exercises...

    I hope that someone would be kind enough to help me out! Thank you!
     
    Last edited: Oct 1, 2007
  2. jcsd
  3. Sep 30, 2007 #2

    HallsofIvy

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    Use the triangle inequality but the other way around.
    Since |y- x|= |x- y|< r2, |x|< |x-y|+ |y|, |x|< r2+ |y|, |x|< |x|-1+ |y|, 1< |y|.


    Well, only the obvious. Since you are told to use the Bolzano-Weierstrasse theorem, Let {xn} be a sequence of points in S. Prove that it has a convergent subsequence! You might want to break it into parts: If there happen to be an infinite number of points on a single one of the line segments Li, can you prove that has a convergent subsequence? (Is each Li closed and bounded?) If not then there must be an infinite subsequnce on {ai} with each ai on a different Li. Can you prove that that sequence converges to 0?
     
  4. Sep 30, 2007 #3
    2) I actually have a worked example to a similar question, and they did the same as you suggested, but I don't understand the difference between the following 2 cases and I am not sure why we need 2 cases...can you please explain more on this part?

    Case 1: If there happen to be an infinite number of points on a single one of the line segments L

    Case 2: If not then there must be an infinite subsequnce on {ai} with each ai on a different Li

    I have a test tomorrow and I am still stuck on these kinds of questions...

    Thanks a lot for your help!
     
    Last edited: Sep 30, 2007
  5. Sep 30, 2007 #4

    HallsofIvy

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    Yes, that's true. And what is their limit?

    No, I didn't say that. There must be an infinite sequence {ai} on ONE Li. What is its limit?

     
  6. Sep 30, 2007 #5
    Sorry, for question #2, I am completely lost now...

    Can anyone explain this question from scratch and provide the steps to solve this problem? What theorems do I need to use in these steps?

    Thanks!
     
  7. Oct 1, 2007 #6
    I am sure someone here knows how to solve this problem. Please help me...
     
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