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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. – Ben Millwood Jul 4 '12 at 18:18
Notation: the double arrow should be a single arrow, and $0$ should be $\{0\}$. $$f:X\to\Bbb C-\{0\}$$ or $$f:X\to\Bbb C\setminus\{0\}\;.$$ – Brian M. Scott Jul 4 '12 at 18:20
Shouldn't this follow from compactness alone, or a I missing something? – Alex Becker Jul 4 '12 at 18:20
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 properties – Alina Jul 4 '12 at 18:22
@Alina: You don't need to. – tomasz Jul 4 '12 at 18:31

the composite $X \to \mathbb{C} \setminus 0 \to \mathbb{R}_{> 0}$ given by first applying $f$ then the norm of a vector is a continuous map. Since $X$ is compact so is the image of this map as a subset of $\mathbb{R}_{>0}.$ Moreover by assumption on $X$ this set is connected. Connected compact subsets of $\mathbb{R}_{>0}$ are closed intervals. Then the claim follows.

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You don't really need connectedness. – tomasz Jul 4 '12 at 18:30
yes. still it works out :) – mland Jul 4 '12 at 18:33
that proof is very nice, thank you @mland, I made everything overcomplicated I guess because of the given properties. – Alina Jul 4 '12 at 18:53
Thanks, No Problem :) – mland Jul 5 '12 at 9:10

Define the function $g: X \to \mathbb{R} $ by $g(x) = |f(x)|$, which is continuous. Since X is compact, the result follows by the Extreme Value Theorem.

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Let $g(x)=|f(x)|$, observe that the complex norm is a continuous function from $\mathbb C$ into $\mathbb R$, therefore $g\colon X\to\mathbb R$ is continuous.

Since $X$ is compact and connected the image of $g$ is compact and connected. All connected subsets of $\mathbb R$ are intervals (open, closed, or half-open, half-closed); and all compact subsets of $\mathbb R$ are closed and bounded (Heine-Borel theorem).

Therefore the image of $g$ is an interval of the form $[a,b]$. Let $x_0,x_1\in X$ such that $g(x)=a$ and $g(x_1)=b$.

(Note that the connectedness of $X$ is not really needed, because compact subsets of $\mathbb R$ are closed and bounded, and thus have minimum and maximum.)

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One could mention Heine-Borel by name... – Rudy the Reindeer Jul 5 '12 at 10:51
@Matt: One could also mention other names. For example Cantor's theorem that a continuous function from a compact metric space into a metric space is uniformly continuous, and perhaps replace the metrizability of the domain by some generalized property like ultrafilters or so. – Asaf Karagila Jul 5 '12 at 10:54
No since we're not trying to confuse the OP. We're trying to help them. – Rudy the Reindeer Jul 5 '12 at 11:12
@Matt: Oh, right. :-P – Asaf Karagila Jul 5 '12 at 11:21
: ) ${}{}{}{}{}$ – Rudy the Reindeer Jul 5 '12 at 11:22

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