Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $X$.

share|improve this question
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

3 Answers 3

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.

share|improve this answer
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.

share|improve this answer

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.)

share|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.