Existence of such points in compact and connected  topological space $X$ 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 $X$.
 A: 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.
A: 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.)
A: 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.
