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Math Help - cts fnc on a closed interval

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    cts fnc on a closed interval

    suppose \ that  \ f:[a.b] \to R \ is \ continuous \ with  \ f(x)>0 \ , \forall x \in \ [a,b]  . prove \ that \ \exists m>0 \ such \ that \ f(x) \geq m , \forall x \in [a,b]
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    Quote Originally Posted by flower3 View Post
    suppose \ that  \ f:[a.b] \to R \ is \ continuous \ with  \ f(x)>0 \ , \forall x \in \ [a,b]  . prove \ that \ \exists m>0 \ such \ that \ f(x) \geq m , \forall x \in [a,b]
    Use the high-point/low-point theorem.
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    Quote Originally Posted by flower3 View Post
    suppose \ that \ f:[a.b] \to R \ is \ continuous \ with \ f(x)>0 \ , \forall x \in \ [a,b] . prove \ that \ \exists m>0 \ such \ that \ f(x) \geq m , \forall x \in [a,b]

    Supose not; then for any n\in\mathbb{N}\,\,\,\exists\,x_n\in [a,b]\,\,\,s.t.\,\,\,f(x_n)<\frac{1}{n},

    Now use the continuity of the function on the sequence \{x_n\} to reach a contradiction...

    Tonio
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by flower3 View Post
    suppose \ that \ f:[a.b] \to R \ is \ continuous \ with \ f(x)>0 \ , \forall x \in \ [a,b] . prove \ that \ \exists m>0 \ such \ that \ f(x) \geq m , \forall x \in [a,b]
    Accidental post.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by flower3 View Post
    suppose \ that \ f:[a.b] \to R \ is \ continuous \ with \ f(x)>0 \ , \forall x \in \ [a,b] . prove \ that \ \exists m>0 \ such \ that \ f(x) \geq m , \forall x \in [a,b]
    Alternatively to tonio's and most likely what Plato was hinting at. Since [a,b] is closed and bounded (compact) and f is continuous we have it that f\left([a,b]\right) is closed and bounded (compact). Thus, \inf\text{ }f\left([a,b]\right)=\alpha exists and is in f\left([a,b]\right). Now since \alpha\in f\left([a,b]\right) it follows that \alpha>0, but by the Archimedean principle we know there exists some n\in\mathbb{N} such that \frac{1}{n}<\alpha, and the conclusion follows.
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