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Basic topology question

  1. Aug 28, 2014 #1
    Assume ##|X| > \rho## , let ##r = |X| - \rho##

    Now I am trying to show that ##B(r,x)\subseteq S^c##

    This should be a simple question, but I am struggling trying to find the right inequlity.

    Attempt:

    let ##y## be a point in ##B(r,x)##.

    I know that ##|x - y| < r##.

    I have to somehow show that ##|y| > \rho##

    this is where my argument falls apart:

    ##|y| \leq |y-x| + |x|< r + \rho## (by triangle inequality)

    but this doesn't show that ##|y| > \rho##

    what am I missing?
     
  2. jcsd
  3. Aug 28, 2014 #2

    HallsofIvy

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    It's awfully hard to figure out what you are trying to say here.

    Is "X" a number and |X| its absolute value (as opposed to a vector and its length)? If a vector or point, in R2, R3, or Rn?

    Okay, Is "x" the same as "X", above? I presume that B(r, x) is the ball of radius r centered on x (in some metric space) but what is Sc? I would guess "the compliment of set S" but what is S?

     
  4. Aug 28, 2014 #3




    I apologize my writing of the question is very messy. You are right on both counts. X is a vector in Rn, Sc is the complement of S, and S is a subset of Rn. Sorry for making it messy.
     
  5. Aug 28, 2014 #4

    HallsofIvy

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    Then your question makes no sense. You are given that [itex]|x|> \rho[/itex], and you want to prove that [itex]B(r, x)\subset S^c[/itex], the complement of set S? But what is "S". You don't mention it in the hypotheses.
     
  6. Aug 29, 2014 #5

    Sorry. S is B([itex]\rho[/itex], 0) , that is the ball of radius [itex]\rho[/itex] about the origin.
     
  7. Aug 29, 2014 #6

    Fredrik

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    OK, you want to show that if ##0<\rho<\|x\|## and ##r=\|x\|-\rho##, then ##B(x,r)\subseteq B(0,\rho)^c##. Let ##y\in B(x,r)## be arbitrary. You want to show that ##\|y\|>\rho##. I suggest this as the first step:
    $$\|y\|\geq\|x\|-\|y-x\|.$$
     
  8. Aug 29, 2014 #7


    Ok using: $$\|y\|\geq\|x\|-\|y-x\|.$$

    I was trying this before but didn't feel it would be valid. Now since ##r=\|x\|-\rho##, I can then say

    ## r + \rho = \|x\|##

    now can I say that:

    ##\|y\|\geq\|x\|-\|y-x\|\geq r + \rho = \|x\| - r##

    which wold reduce to ##\|y\|\geq\rho ##
     
  9. Aug 29, 2014 #8

    Fredrik

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    Are you familiar with the triangle inequality in the form ##\|x+y\|\geq \|x\|-\|y\|##? (This can be derived from the usual version). Write ##\|y\|=\|(y-x)+x\|## and then use this.

    This should be ##\|y\|\geq\|x\|-\|y-x\| > r+\rho-r##. (Because ##\|x\|=r+\rho## and ##\|y-x\|<r##).
     
  10. Aug 29, 2014 #9
    Yea I just copied the code wrong for that, but no i was not aware of the triangle inequality in that form. I'm gping to go derive it now. Thanks. If I have an issue I will ask for assistence
     
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