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Thread: inf(S)

  1. #1
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    inf(S)

    ) Let g : [0; 1] ! R be a continuous function. Let S be a nonempty subset of [0; 1]
    and suppose that g(x) = 3 for all x 2 S. Let a = inf(S). Show that g(a) = 3
    Last edited by mr fantastic; Nov 27th 2011 at 10:15 AM. Reason: Re-titled.
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  2. #2
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    Re: need question help

    I'm not sure that I understand your question, but if S is a closed set then

    $\displaystyle a=\inf(S)=\min(S)$

    and since $\displaystyle g(x)=3$ for all $\displaystyle x\in S$, we conclude that $\displaystyle g(a)=3$. If $\displaystyle S$ is an open set you should find a sequence of numbers that converges to $\displaystyle a$ and use the continuity property to prove that $\displaystyle \lim\limits_{x_n\rightarrow a} g(x_n)=3$ (follow this link)
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  3. #3
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    Re: need question help

    Quote Originally Posted by uasac View Post
    I'm not sure that I understand your question, but if S is a closed set then $\displaystyle a=\inf(S)=\min(S)$
    and since $\displaystyle g(x)=3$ for all $\displaystyle x\in S$, we conclude that $\displaystyle g(a)=3$. If $\displaystyle S$ is an open set you should find a sequence of numbers that converges to $\displaystyle a$ and use the continuity property to prove that $\displaystyle \lim\limits_{x_n\rightarrow a} g(x_n)=3$ (follow this link)
    It is not the case that $\displaystyle S$ is either open or closed.
    It is the case that $\displaystyle a\in S\text{ or }a\notin S$.
    Now this argument works.
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  4. #4
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    Re: need question help

    If $\displaystyle S$ is open then $\displaystyle a\notin S$.
    If $\displaystyle S$ is closed then $\displaystyle a\in S$.
    I think...
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  5. #5
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    Re: need question help

    Quote Originally Posted by uasac View Post
    If $\displaystyle S$ is open then $\displaystyle a\notin S$.
    If $\displaystyle S$ is closed then $\displaystyle a\in S$.
    I think...
    What you have written there is correct.
    BUT the set $\displaystyle S$ may be neither open nor closed.
    It could be that $\displaystyle S=\{0.1\}\cup(0.5,1].$
    That set is neither open or closed.
    However, $\displaystyle \inf(S)\in S.$

    It could be that $\displaystyle S=(0.1,0.5]\}\cup\{1\}$
    That set is neither open or closed.
    However, $\displaystyle \inf(S)\notin S.$
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  6. #6
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    Re: need question help

    Right... thanks for the correction.
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