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Proff of half infinite intervals through set theory

  1. Nov 18, 2009 #1
    Problem:
    We define half infinite intervals as follows:
    (a, [tex]\infty[/tex]) = {x[tex]\in[/tex] R | x>a};
    [a, [tex]\infty[/tex]) = {x[tex]\in[/tex] R | x[tex]\geq[/tex]a};

    Prove that:
    (i) (a, [tex]\infty[/tex]) [tex]\subseteq[/tex] [b, [tex]\infty[/tex]) [tex]\Leftrightarrow[/tex] a[tex]\geq[/tex]b,
    (ii) [a, [tex]\infty[/tex]) [tex]\subseteq[/tex] (b, [tex]\infty[/tex]) [tex]\Leftrightarrow[/tex] a>b.

    I've got pretty much no idea how to do this. Then again I've been struggling at this for a couple hours and my mind doesn't work particularly well at 2:25 am PST. Help would be greatly appreciated on this.

    Thanks.
     
  2. jcsd
  3. Nov 18, 2009 #2

    HallsofIvy

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    You prove that [itex]A\subseteq B[/itex] by proving that any member of A is a member of B.
    If [itex]a\geq b[/itex], then for any member, x, of [itex](a, \infty)[/itex], [itex]x> a\ge b[/itex] so x> b and therefore [itex]x\in [b, \infty)[/itex].

    The converse is a litlle harder because we are looking at [itex](a, \infty)[/itex] rather than [itex][a,\infty)[/itex], for all [itex]\epsilon> 0[/itex], [itex]a+ \epsilon> a[itex] so [itex]a+ \epsilon\in (a, \infty)[/itex]. If [itex](a,\infty)\subseteq [b, \infty)[/itex] and so [itex]a+\epsilon\ge b[/itex]. Taking the limit as [itex]\epsilon[/itex] goes to 0, [itex]a\le b[/itex].

    (ii) is actually simpler.
     
  4. Nov 18, 2009 #3
    much appreciated. cleared things up for me.
     
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