Published Jul 01, 2021
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Theorem 4.4.1 (Preservation of Compact Sets) Let \(f \to \mathbb{R}\) be continuous on \(A\). If \(K \subseteq A\) is compact, then \(f(K)\) is compact as well.
Theorem 4.4.2 (Extreme Value Theorem). If \(f: K \to \mathbb{R}\) is continuous on a compact set \(K \subseteq \mathbb{R}\), then \(f\) attains a maximum and minimum value. In other words, there exists \(x_0, x_1 \in K\) such that \(f(x_0) \leqslant f(x) \leqslant f(x_1)\) for all \(x \in K\)
Definition 4.4.4 (Uniform Continuity). A function \(f: A \to R\) is uniformly continuous on \(A\) if for every \(\epsilon > 0\) there exists a \(\delta > 0\) such that for all \(x,y \in A\), \(|x - y| < \delta\) implies \(|f(x) - f(y)| < \epsilon\).
Theorem 4.4.5 (Sequential Criterion for Absence of Uniform Continuity). A function \(f: A \to \mathbb{R}\) fails to be uniformly continuous on \(A\) if and only if there exists a particular \(\epsilon_0 > 0\) and two sequences \((x_n)\) and \((y_n)\) in \(A\) satisfying $$ |x_n - y_n| \to 0 \text{ but } |f(x_n) - f(y_n)| \geqslant \epsilon_0 $$
Theorem 4.4.7 (Uniform Continuity on Compact Sets). A function that is continuous on a compact set \(K\) is uniformly continuous on \(K\)