$\displaystyle suppose \ that \ f:[a.b] \to R \ is \ continuous \ with \ f(x)>0 \ , \forall x \in \ [a,b] $ $\displaystyle . prove \ that \ \exists m>0 \ such \ that \ f(x) \geq m , \forall x \in [a,b] $

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- Dec 19th 2009, 10:09 AMflower3cts fnc on a closed interval
$\displaystyle suppose \ that \ f:[a.b] \to R \ is \ continuous \ with \ f(x)>0 \ , \forall x \in \ [a,b] $ $\displaystyle . prove \ that \ \exists m>0 \ such \ that \ f(x) \geq m , \forall x \in [a,b] $

- Dec 19th 2009, 10:21 AMPlato
- Dec 19th 2009, 10:24 AMtonio
- Dec 19th 2009, 09:31 PMDrexel28
- Dec 19th 2009, 09:37 PMDrexel28
Alternatively to

**tonio**'s and most likely what**Plato**was hinting at. Since $\displaystyle [a,b]$ is closed and bounded (compact) and $\displaystyle f$ is continuous we have it that $\displaystyle f\left([a,b]\right)$ is closed and bounded (compact). Thus, $\displaystyle \inf\text{ }f\left([a,b]\right)=\alpha$ exists and is in $\displaystyle f\left([a,b]\right)$. Now since $\displaystyle \alpha\in f\left([a,b]\right)$ it follows that $\displaystyle \alpha>0$, but by the Archimedean principle we know there exists some $\displaystyle n\in\mathbb{N}$ such that $\displaystyle \frac{1}{n}<\alpha$, and the conclusion follows.