Extreme Value Theorem

Last updated Nov 6, 2022

Let $X$ be a Topological Spaces and let $K \subset X$ be Nonempty and Compact. Let $Y$ be a Total Ordering endowed with the Order Topology. Let $f: X \to Y$ be a Continuous Function. Then $\sup\limits f(K)$ and $\inf\limits f(K)$ both exist and are achieved by some resp. $x \in K$ (or, in short, $f(K)$ is tightly bounded).

Because Continuous Functions Preserve Compactness, we know $f(K)$ is Compact Since $K$ is Nonempty, we know $f(K)$ is Nonempty. A Nonempty Set is Compact in the Order Topology iff it is Tightly Bounded and Complete, so both $\sup\limits f(K) \in f(K)$ and $\inf\limits f(K) \in f(K)$. This means precisely that there exists $x,y \in K$ so that $f(x) = \sup\limits f(K)$ and $f(y) = \sup\limits f(K)$. $\blacksquare$

Let $X$ be a Topological Space and let $K \subset X$ be Compact. Let $f: X \to \mathbb{R}$ be a Continuous Function. Then $\sup\limits f(K)$ and $\inf\limits f(K)$ both exist and are achieved by some resp. $x \in K$.

Follows from Statement (1) after noting that the standard topology on the Real Numbers is the Order Topology with the natural ordering. $\square$