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Proof

(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.
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References

Bibliography

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