Maximum and Minimum Theorem in Metric Spaces 📂MetricSpace

Maximum and Minimum Theorem in Metric Spaces

Theorem

Let $X$ be a compact metric space, and let $f : X \to \mathbb{R}$ be continuous. Then, it is as follows.

$M = \sup \limits_{x\in X} f(x),\quad m=\inf \limits_{x \in X}f(x)$

Then,

$M=f(p),\quad m=f(q)$

there exists a $q,p\in X$ that satisfies this. In other words: for every $x$ ,

$f(q)\le f(x) \le f(p)$

there exists a $q,p \in X$ that satisfies this. This is called the extreme value theorem.

Explanation

The compactness condition is necessary.

This theorem guarantees that $f(X)$ includes the maximum and minimum values of $f$ . Without any conditions, by the definition of supremum and infimum, there is no guarantee that $M$ , $m$ are included in $f(X)$ , but the assumption that $X$ is compact and $f$ is continuous makes $M,m\in f(X)$ valid.

Proof

Since $f$ is continuous in a compact space, $f(X)$ is compact. According to the equivalent condition for compactness in Euclidean space, $f(X)$ is a closed and bounded set of real numbers.

Auxiliary theorem
Let $E$ be a non-empty set of real numbers and bounded above. And let $y=\sup E$ . Then, $y \in \overline{E}$ . Furthermore, if $E$ is closed, then $y \in E$ .

Then, the proof is complete by the auxiliary theorem.

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