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Math Help - Proving (-infinity, 0) is open

  1. #1
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    Proving (-infinity, 0) is open

    Hi, I've done most of the proof(I think), but I'm lost anyone willing to finish it off for me? I've spent a while on it and i'm stuck.

    Let x be an element of (-infinity, 0) this implies x < 0 chose an r such that abs(r) < abs(x) let y be an element of (x -r, x+r).
    This implies x-r<y<x+r which implies 0 - x + x > x - r < y now what exactly do I do to show y < 0 which would be the end goal, right?
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  2. #2
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    Re: Proving (-infinity, 0) is open

    Quote Originally Posted by glambeth View Post
    Hi, I've done most of the proof(I think), but I'm lost anyone willing to finish it off for me? I've spent a while on it and i'm stuck.
    Let x be an element of (-infinity, 0) this implies x < 0 chose an r such that abs(r) < abs(x) let y be an element of (x -r, x+r).
    This implies x-r<y<x+r which implies 0 - x + x > x - r < y now what exactly do I do to show y < 0 which would be the end goal, right?
    If x\in (-\infty,0) then x\in (2x,0)\subset (-\infty,0) . (2x,0) is an open set containing x.

    Does that work?
    Thanks from topsquark
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  3. #3
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    Re: Proving (-infinity, 0) is open

    Hm that makes sense, but how would I be able to get to that stage?
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  4. #4
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    Re: Proving (-infinity, 0) is open

    What you did was perfectly good- the "r-neighborhood" of x, (x-r, x+ r) is a subset of (-\infty, 0) so x is an interior point of (-\infty, 0). Since x could be any point in the set, the set is open.
    Last edited by Plato; September 29th 2013 at 05:52 PM.
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  5. #5
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    Re: Proving (-infinity, 0) is open

    Hm, I think my part is wrong though. Looking at it i say choose an r such that abs(r) < abs(x) but then i say x - r < y < x + r so i end up saying 0 < y < 2x but isn't this false? Given y should be less than 0?
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  6. #6
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    Re: Proving (-infinity, 0) is open

    If |r|<|x|, it is possible that r<0, which you don't want. You want 0<r<|x|. Now, (x-r,x+r) is correct. This is because x<0, so x-r<0. Since 0<r<|x|, x+r = -|x|+r<0. So, you have both endpoints less than zero.
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