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Open Set Proofs

  1. Sep 24, 2009 #1
    1. The problem statement, all variables and given/known data

    Using the definition of an open set, prove that the subset of S ={(x, y)∈ R2 | 0 <x< 1, 1< y<2} the Euclidean space R2 is open.


    3. The attempt at a solution

    The definition I learnt is a set S [tex]\subset[/tex] Rn is said to be open in Rn if for all x[tex]\in[/tex] S, [tex]\exists[/tex] r greater than 0, such that every point y [tex]\in[/tex] Rn satisfying ||x-y|| < r also belongs to S.
    So, what I think is choose r =x for all x ∈ S, there exists r greater than 0, such that if |x1 - x | < r then B(r,x)[tex]\subset[/tex] S, S is open.

    but I think it is too simple, I really have no idea how to prove a set open with all the inequalities and functions going on..
     
    Last edited: Sep 24, 2009
  2. jcsd
  3. Sep 25, 2009 #2

    lanedance

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    I think you need to show that r exists, as you defined, for any point in your set....

    so say you take
    [tex] \textbf{s} = (s,t) \in S = \left\{ (x, y) \in \mathbb{R}^2 | 0 < x < 1, 1< y<2 \right\} [/tex]

    now find r, such that for any p = (p,q) with |p-s| < r, then p is also in S

    if you can demonstrate this is true for any arbitrary point s of S, then you have shown it is true for every point in S, and so S is open
     
    Last edited: Sep 25, 2009
  4. Sep 26, 2009 #3
    I don't think what lanedance said about the part where if |p-s| < r, then p is also in S, is true. That's what you're trying to prove. So, you can't assume that. Well, it's true that to show the set S is open, you need to find r such that the open ball about s = (s,t) with radius r is totally contained in S.
    One instance of r that I have in mind right now is by letting r = min{s, 1-s, t-1, 2-t} >0. Then, once you've define your r, you need to prove that the open ball is contained in S. To show this, let p = (p1,p2) be an arbitrary point in B(s, r). So, this is implies that d(p, s) < r . Now, with that information, you need to prove that p also belongs to the set S. To do this, you've got to show that 0<p1<1 and 1<p2<2. Once you show this, it implies that p belongs to S by definition. Therefore, the open ball B(s, r) is contained in S.
     
  5. Sep 26, 2009 #4

    lanedance

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    yes it is, that is the definition of open in terms of open balls, which seems to be exactly what you have done below

     
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