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Proving inf(ST) = inf(S)*inf(T)

  1. Jul 17, 2011 #1
    1. The problem statement, all variables and given/known data

    Sets A and B are sets of positive real numbers. Define C = {st| s \in S and t \in T}
    Prove inf(C) = inf(S)*inf(T)

    3. The attempt at a solution

    so i'm trying to prove inf(C) <= inf(S)*inf(T) and inf(C) >= inf(S)*inf(T).

    i'll use e as epsilon. epsilon is positive

    By definition there is an s in S such that: s < inf S + e. There is a t in T such that: t < inf T + e. Additionally, inf(C) <= st for all s in S and t in T by definition.

    st < (inf S + e)(inf T + e) = (inf S)(inf T) + (inf S)*e + (inf T)*e + e^2

    inf C <= st <= (inf S)(inf T) < inf S)(inf T) + (inf S)*e + (inf T)*e + e^2
    (NOTE: i'm not sure about the middle inequality: "st <= (inf S)(inf T)" )

    is this a correct way to go about this? I'm also not sure about proving the other direction
     
  2. jcsd
  3. Jul 17, 2011 #2

    micromass

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    No, the middle inequality is indeed not correct. Try it like this:

    [tex]inf(C)\leq st<inf(S)inf(T)+inf(S)e+inf(T)e+e^2[/tex]

    Because e is arbitrary, we can let e go to 0, thus

    [tex]inf(C)<inf(S)\inf(T)[/tex]

    For the other inequality, take [itex]st<inf(C)+e[/itex] and do something with it.
     
  4. Jul 17, 2011 #3

    Hurkyl

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    You're right to be unsure. You have
    • [itex]\inf C \leq st[/itex]
    • [itex]st < (\inf S)(\inf T) + (\inf S) \epsilon + (\inf T) \epsilon + \epsilon^2[/itex]
    So you just put the two together:
    [itex]\inf C \leq st< (\inf S)(\inf T) + (\inf S) \epsilon + (\inf T) \epsilon + \epsilon^2[/itex]​
    well, what you really care about is just transitivity:
    [itex]\inf C < (\inf S)(\inf T) + (\inf S) \epsilon + (\inf T) \epsilon + \epsilon^2[/itex]​
    And then invoke what you can about the fact that this is true for every positive real number [itex]\epsilon[/itex].

    (aside: micromass forgot that < turns into [itex]\leq[/itex] when doing limits)
     
  5. Jul 17, 2011 #4

    micromass

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    Oh my, I'm still not fully awake :frown:
     
  6. Jul 17, 2011 #5
    For the other direction I'm guessing it's:

    There is an st in C such that st < inf C + e.
    By definition, inf S <= s for all s and inf T <= t for all t. So (inf S)(inf T) <= st.

    So we have (inf C + e) > st => (inf S)(inf T). so (inf C + e) > (inf S)(inf T). (I need this to be a => though)

    I'm not quite sure what you mean by "invoking what you know about e". Does this mean that the ">" becomes a "=>". See below for context

     
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