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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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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.2012-07-04
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    yes. still it works out :)2012-07-04
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    that proof is very nice, thank you @mland, I made everything overcomplicated I guess because of the given properties.2012-07-04
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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...2012-07-05
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    @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.2012-07-05
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    No since we're not trying to confuse the OP. We're trying to help them.2012-07-05
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    @Matt: Oh, right. :-P2012-07-05
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    : ) ${}{}{}{}{}$2012-07-05
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    Perhaps a nitpick, but I always think of the Heine--Borel theorem as saying that closed and bounded subsets of $\mathbb R^n$ are compact, not the converse (which applies in every metric space).2012-07-06
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    @Jonas: I was always under the impression that the theorem was "iff".2012-07-06
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    @Jonas: Also to be correct, in the general setting of a metric space one has to require a set to be totally bounded, not just bounded. The notions coincide for $\mathbb R^n$, as you probably know.2012-07-06
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    @Asaf: I'm just saying that a compact subset of a metric space is closed and bounded, and that is all that is used here, not the special fact about $\mathbb R^n$ that the converse holds, i.e. Heine--Borel or one half of it if you prefer. It is true that total boundedness also holds, and conversely if a metric space is complete and totally bounded then it is compact (at least assuming choice), but none of that seems very helpful here.2012-07-08
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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.