Let $f$ be usc and less than +$\infty$ on a compact set $E$ , show that $f$ assumes its maximum on $E$.
First, I show $f$ is bounded above on $E$
Since $f$ is usc on $E$
{$f<a_k$} is relative open on $E$
{$f<a_k$} is open covering of $E$
since $E$ is compact
so exist $a_1,...,a_n$ such that $$E\subseteq \bigcup_{k=1}^{n}\{f<a_k\}$$
Take $N=max\{a_1,...,a_n\}$
then $f$ is dounded above by $N$ on $E$.
Second,I use the fact $f$ is bounded above
Take $M=\sup f(x)$,$x\in E$
It is clearly that $M\ge f(x)$ , for $x\in E$
but I have no idea about $M\le \sup f(x)$
Can you give me some suggestion,thank you!!!
Your proof of the boundedness of $f$ is correct, but you don't need to show $M \le \sup f$, since you have defined $M$ as the supremum of $f$ on $E$. You just need to prove that the maximum is actually attained by $f$.
By the definition of $M$, for every positive integer $n$ the set $C_n = \{x \in E \mid f(x) \ge M - \tfrac{1}{n}\}$ is non-empty and by upper semicontinuity closed. Since $E$ is compact, this is a decreasing sequence of nonempty compact sets and Cantor's intersection theorem tells us that the intersection $C = \bigcap \limits_{n = 1}^\infty C_n$ is nonempty. Let $x_0 \in C$ be arbitrary, then $f(x_0) \ge M - \frac{1}{n}$ holds for every positive integer $n$. Taking the limit for $n \to \infty$ we get $f(x_0) \ge M$. Since $f(x_0) \le M$ holds trivially, we have proven $f(x_0) = M$.