The Extreme Value Theorem for Cts. Fns. on Comp. Sets of Met. Sps.

The Extreme Value Theorem for Continuous Functions on Compact Sets of Metric Spaces

Recall from the Continuous Functions on Compact Sets of Metric Spaces page that if $(S, d_S)$ and $(T, d_T)$ are metric spaces, $X \subseteq S$ is a compact subset of $S$, and $f : S \to T$ is continuous, then the image $f(X)$ is a compact subset of $T$.

Furthermore, we also noted that since $f(X)$ is a compact subset of $T$ that then $f(X)$ is also closed in $T$, bounded in $T$, and every infinite subset of $f(X)$ has an accumulation point.

We will now apply this theorem to state a more generalized Extreme Value Theorem for a metric space that maps into $\mathbb{R}$ with the usual Euclidean metric.

Theorem 1 (The Extreme Value Theorem): Let $(S, d_S)$ and $(\mathbb{R}, d)$ be metric spaces where $d$ is the usual Euclidean metric defined for all $x, y \in \mathbb{R}$ by $d(x, y) = \mid x - y \mid$. Also let $X \subseteq S$ be a compact subset of $S$, and $f : S \to T$ be continuous on all of $X$. Then $f(X)$ is closed and bounded in $T$ and $f$ achieves its supremum and infimum on $X$, that is, there exists $p, q \in X$ such that $f(p) = \sup \{ f(x) : x \in X \}$ and $f(q) = \inf \{ f(x) : x \in X\}$.
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  • Proof: Since $X \subseteq S$ and $f : S \to T$ is continuous on all of $X$ we have by the theorem summarized about that $f(X)$ is a compact subset of $T$ and is also closed and bounded in $T$. We only need to prove that there exists $p, q \in S$ such that $f(p) = \sup \{ f(x) : x \in X \}$ and $f(q) = \inf \{ f(x) : x \in X \}$.
  • Let $L = \sup \{ f(x) : x \in X \}$. Then $L$ is the least upper bound of $f$ as $x$ ranges through $X$. For all $x \in X$ we then have that $f(x) \leq L$. Furthermore, since $L$ is the least upper bound, we have that for all $\epsilon > 0$ that there exists an $x_{\epsilon} \in X$ such that $L - \epsilon \leq f(x_{\epsilon})$ (if not, then $L - \epsilon$ would be an upper bound of $f(x)$ that is less than $L$ which would be a contradiction. Hence:
(1)
\begin{align} \quad L - \epsilon \leq x_{\epsilon} \leq L \end{align}
  • So, for all $\epsilon > 0$ we have that $B(L, \epsilon) \cap f(X) \neq \emptyset$, so $L$ is an adherent point of $f(X)$. But $f(X)$ is closed in $T$ and contains all of its adherent points, so there exists a $p \in X$ such that:
(2)
\begin{align} \quad f(p) = L = \sup \{ f(x) : x \in X \} \end{align}
  • Similarly let $l = \inf \{ f(x) : x \in X \}$. Then $l$ is the greatest lower bound of $f$ as $x$ ranges through $X$. For all $x \in X$ we then have that $l \leq f(x)$. Furthermore, since $l$ is the greatest lower bound we have that for all $\epsilon > 0$ that there exists an $x_{\epsilon} \in X$ such that $f(x) \leq l + \epsilon$ (if not, then $l + \epsilon$ would be a lower bound of $f(x)$ that is greater than $l$ which would be a contradiction to $l = \inf \{ f(x) : x \in X \}$. Hence:
(3)
\begin{align} \quad l \leq f(x_{\epsilon}) \leq l + \epsilon \end{align}
  • So for all $\epsilon > 0$ we have that $B(l, \epsilon) \cap f(X) \neq \emptyset$, so $l$ is an adherent point of $f(X)$. But $f(X)$ is closed in $T$ and contains all of its adherent points, so there exists a $q \in X$ such that:
(4)
\begin{align} \quad f(q) = l = \inf \{ f(x) : x \in X \} \quad \blacksquare \end{align}
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