(related to Corollary: Continuous Functions Mapping Compact Domains to Real Numbers are Bounded)

- Let $X$ be a metric spaces.
- Let $D\subset X$ be a compact subset.
- Let $f:D\mapsto \mathbb R$ be a continuous function mapping the domain $D$ to the real numbers $\mathbb R$.
- We have to show that $f$ is bounded.
- According to the corresponding theorem, there are points \(p,q\in X\), for which the function $f$ takes the maximum and the maximum value, i.e. $f(p)=\max(f(D))$ and $f(q)=\min(f(D))$.
- Take some real numbers $P,Q\in\mathbb R$ with $$Q\le f(q)\le f(x)\le f(p) \le P$$ for all \(x\in D\).
- It follows that $P$ and $Q$ are the upper and the lower bounds of the image $f(D)$, i.e. we have $$ Q\le f(x) \le P$$ for all $f(x)\in f(D)$.
- It follows from the definition of bounded functions that $f$ is bounded.∎

**Forster Otto**: "Analysis 2, Differentialrechnung im \(\mathbb R^n\), Gewöhnliche Differentialgleichungen", Vieweg Studium, 1984