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Math Help - Basic Proof

  1. #1
    Super Member Aryth's Avatar
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    Basic Proof

    Let x and y be distinct real numbers. Prove that there is a neighborhood P of x and a neighborhood Q of y such that P \cap Q = \emptyset.

    The book left this proof as an exercise and the instructor said nothing about it. I'm just curious as to what the proof might be.
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    Re: Basic Proof

    Have you considered a neighborhood smaller than |P-Q|?
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    Super Member Aryth's Avatar
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    Re: Basic Proof

    I have tried to consider your suggestion, and it looks similar to the epsilon we chose when proving that a sequence can't converge to two numbers unless they were equal. However, I'm not quite sure what I'm supposed to do with that... We just assumed the OP was true and moved on with it.

    P.S. I don't have much experience proving things so it's taking me awhile to think things through.
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    Re: Basic Proof

    Quote Originally Posted by Aryth View Post
    I have tried to consider your suggestion, and it looks similar to the epsilon we chose when proving that a sequence can't converge to two numbers unless they were equal. However, I'm not quite sure what I'm supposed to do with that... We just assumed the OP was true and moved on with it.
    P.S. I don't have much experience proving things so it's taking me awhile to think things through.
    Let \delta=\frac{|x-y|}{4}>0.
    Let P=(x-\delta,x+\delta)~\&~P=(y-\delta,y+\delta).
    Show that P\cap Q=\emptyset .
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    Super Member Aryth's Avatar
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    Re: Basic Proof

    Just curious, why did you pick that as \delta?
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    Re: Basic Proof

    Quote Originally Posted by Aryth View Post
    Just curious, why did you pick that as \delta?
    Because it works.
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    Super Member Aryth's Avatar
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    Re: Basic Proof

    That's basically the same thing my instructor said. To me it just seems random, but it does work.

    I appreciate your help.
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    Re: Basic Proof

    The point is that if a specific \delta works, then so does any positive number less than \delta.

    TKHunny suggested \deltas= \frac{|P- Q|}{2} which is the largest number that would work so that N_P(\delta) and N_Q(\delta) are disjoint because \delta+ \delta= \frac{|P-Q|}{2}+ \frac{|P-Q|}{2}= |P- Q|. (Draw a picture to see why.) Plato just chose to pick a smaller number. \delta= \frac{|P-Q|}{3} would also work.

    And, that is using the same \delta for both sets, which is not necessary. You could also use, say, \frac{|P-Q|}{5} for one open set and \frac{|P-Q|}{3} for the other- just use two numbers that add up to less than |P-Q|.
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    Super Member Aryth's Avatar
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    Re: Basic Proof

    Ah, that makes it clearer. Thank you very much.
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