Proving every continous function reaches maximum on a set A. Let $A$ be a bounded set in $R^{n}$ with the following propety:
Every continous function (over $A$) $F:A \subset\ R^{n}\rightarrow R$, such that $F$ is strictly positive reaches a minimum on $A$.
Prove every continous function (over $A$) $F:A \subset\ R^{n} \rightarrow R$ reaches a maximum on A.
 A: proof:
let $F: A \to \Bbb R$ be any continuous function. Define $G: A \to \Bbb R$ as
$$G(x)= e^{-F(x)}$$
This function is strictly positive and continuous, hence by hypothesis $G$ has a minimum point $x_0 \in A$.
Since $e^{-t}$ is strictly decreasing in $t$, you have that $x_0$ is a maximum point for $F$.
A: We claim that $A$ is compact. To prove this we use (one form of) Tietze Extension Theorem. Consider $A$ as  a metric space in its own right. Suppose $A$ is not compact. Then there is a sequence $\{x_n\}$ in $A$ with no limit point. The set $\{x_n:n \geq 1\}$ is a closed set in $A$. Let $f(x_n)=\frac 1 n $. This is a continuous map from $\{x_n:n \geq 1\}$ into $(0,1)$. By Tietze's theorem there exists a continuous function $F:A\to (0,1)$ such that $F=f$ on $\{x_n:n \geq 1\}$. This is a positive continuous function which has no minimum on $A$. This contradiction proves that $A$ is compact. Hence every continuous real function on $A$ attains its maximum. Note: Tietze Extension Theorem is usually stated for functions from a metric space into $[0,1]$ or into $\mathbb R$. Since $(0,1)$ is homeomorphic to $\mathbb R$ we can also take the range to be $(0,1)$. 
A: If $ A $ is compact then $ f $ is uniformly continuous on $ A $, so there is a partitioning such that for each partition we have $ |max (f)- min (f)| < \epsilon $ for  every $x \in A $  , where $\epsilon  > 0$
