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Thread: Infimum

  1. #1
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    Infimum

    If I use $\displaystyle x\leq\frac{k}{2^n}$ to show E is nonempty and bounded above, then E has a supremum. What would I use to show: If E is nonempty and bounded below, then E has an infimum. Would $\displaystyle x\geq\frac{k}{2^n}$ work? When I tried this the induction part doesn't make sense to me. For example by induction, then, there exists integers $\displaystyle k_n$ least in $\displaystyle A_n$ such that $\displaystyle k_o \leq \frac{k_1}{2} \leq \frac{k_2}{4} \leq ... \leq \frac{k_n}{2^n} \leq ...$ and to me that doesn't seem to make sense for something bounded below.
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  2. #2
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    Will you please just state exactly what you are required to prove?
    Is it, “If a set has a lower bound, then it has an infimum”?
    If so, do you know that if a set has an upper bound them it has a supremum?
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  3. #3
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    Yes... Prove that if E is nonempty and bounded below, then E has an infinum. Sorry about the confusion.
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  4. #4
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    If you know that a set has an upper bound has a supremum, then the proof is simple.
    Consider $\displaystyle F = \left\{ {x:\left( {\forall z \in E} \right)\left[ {x \leqslant z} \right]} \right\}$. That is $\displaystyle F$ is the set of all lower bounds for $\displaystyle E$.
    Then $\displaystyle F$ is a nonempty set bounded above. Therefore $\displaystyle F$ has a supremum, $\displaystyle \gamma$.
    Then, you show that $\displaystyle \gamma = \inf (E)$
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  5. #5
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    How did you "use $\displaystyle k/2^n$" to prove that non-empty set, with upper bound, has a supremum? By using monotone convergence? You can do something similar by multiplying by -1 to make the increasing sequence $\displaystyle k/2^n$ into the decreasing sequence $\displaystyle -k/2^n$. Or do as Plato suggested and use the fact about supremum to prove infimum- again multiply by -1 to reverse the order.
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  6. #6
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    Yes by using the monotone sequence property. This proof I'm working on atm does not specify that the set is bounded above or has a supremum.
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