Maximum Value Theorem proof in a eucledian sapce

Question

I need help with the following proof:Prove Weierstrass Maximum Theorem for the special case in which $(X; dX) $ is the usual Euclidean vector space $(\Bbb{R}^n;d2)$ .

Answer 1

I guess you mean Weierstrass extreme value theorem. A generalization of its case for the reals proposed in Wikipedia states that if $K$ is a compact set and $f:K\to\Bbb R$ is a continuous function, then $f(K)$ is bounded and there exist $p,q\in K$ such that $f(p)=\sup_{x\in K} f(x)$ and $f(q)=\inf_{x\in K} f(x)$. In Wikipedia continuity of the function $f$ is defined in terms of open sets, but it is easy to show that a function $f$ is continuous according to Wikipedia definition iff $f$ is continuous according to the usual definition in terms of $\varepsilon$ and $\delta$. Also recall that by Heine-Borel Theorem, a subset $S$ of $\Bbb R^n$ is compact iff $S$ is closed (that is, for each $x\in \Bbb R^n\setminus S$ there exists $\varepsilon>0$ such that for each $y\in \Bbb R^n$ with $dX(x,y)<\varepsilon$ we have $y\in \Bbb R^n\setminus S$ too) and bounded (that is, for each $\varepsilon>0$ there exists a finite subset $F$ of $S$ such that for each $x\in S$ there exists $y\in F$ such that $dX(x,y)<\varepsilon$).

Therefore we have the following generalization of Weierstrass extreme value theorem.

Theorem If $K\subset\Bbb R^n$ is a non-empty closed and bounded set and $f:K\to\Bbb R$ is a continuous function, then $f(K)$ is bounded and there exist $p,q\in K$ such that $f(p)=\sup_{x\in K} f(x)$ and $f(q)=\inf_{x\in K} f(x)$.

In order to prove the theorem we use the preservation of compactness by continuous maps.

Lemma. If $V, W$ are topological spaces, $F:V\to W$ is a continuous function, and $K\subset V$ is compact, then $f(K)\subset W$ is also compact.

Proof. Let $\mathcal W$ be any open cover of the set $f(K)$. By continuity of the map $f$, a family $\{f^{-1}(U):U\in\mathcal W\}$ is an open cover of the set $K$. Since the set $K$ is compact, it has a finite subcover $\{f^{-1}(U):U\in\mathcal W_0\}$. Then $\mathcal W_0$ is a finite cover of the set $f(K)$. $\square$

Proof of Theorem. By Heine-Borel Theorem, $K$ is compact. By Lemma, $f(K)$ is compact. Again by Heine-Borel Theorem, $f(K)$ is closed and bounded in $\Bbb R$. It is easy to show that $f(K)$ is a segment, that is $f(K)=[\inf_{x\in K} f(x), \sup_{x\in K} f(x)]$. Thus $\inf_{x\in K} f(x)\in f(K)$ and $\sup_{x\in K} f(x)\in f(K)$. $\square$