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Let $X$ be a topological space which is compact and connected.

$f$ is a continuous function such that;

$f : X \to \mathbb{C}-\{0\}$.

Explain why there exists two points $x_0$ and $x_1$ in $X$ such that $|f(x_0)| \le |f(x)| \le |f(x_1)|$ for all $x$ in $X$.

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    You should explain what you've already tried, and whether or not you understand the concepts involved.2012-07-04
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    Notation: the double arrow should be a single arrow, and $0$ should be $\{0\}$. $$f:X\to\Bbb C-\{0\}$$ or $$f:X\to\Bbb C\setminus\{0\}\;.$$2012-07-04
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    Shouldn't this follow from compactness alone, or a I missing something?2012-07-04
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    I've tried to use X's being compact so it means that it is bounded and inf(X) and sup(X) are in X but I have no idea how to use path connected and locally connected properties2012-07-04
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    @Alina: You don't need to.2012-07-04
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    See also this question: [$X$ compact metric space, $f:X\rightarrow\mathbb{R}$ continuous attains max/min](http://math.stackexchange.com/questions/109548/x-compact-metric-space-fx-rightarrow-mathbbr-continuous-attains-max-min) or Corollary 3 in the ProofWiki article [Continuous Image of a Compact Space is Compact](http://www.proofwiki.org/wiki/Continuous_Image_of_a_Compact_Space_is_Compact).2012-07-05

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