3

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 $X$.

Asaf Karagila
  • 405,794
Alina
  • 31

3 Answers3

2

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.

mland
  • 2,682
2

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.)

Asaf Karagila
  • 405,794
  • One could mention Heine-Borel by name... – Rudy the Reindeer Jul 05 '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 05 '12 at 10:54
  • No since we're not trying to confuse the OP. We're trying to help them. – Rudy the Reindeer Jul 05 '12 at 11:12
  • @Matt: Oh, right. :-P – Asaf Karagila Jul 05 '12 at 11:21
  • : ) ${}{}{}{}{}$ – Rudy the Reindeer Jul 05 '12 at 11:22
  • 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). – Jonas Meyer Jul 06 '12 at 06:15
  • @Jonas: I was always under the impression that the theorem was "iff". – Asaf Karagila Jul 06 '12 at 06:57
  • @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. – Asaf Karagila Jul 06 '12 at 20:12
  • @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. – Jonas Meyer Jul 08 '12 at 03:01
2

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.

Ink
  • 5,234