It is said in my textbook that a continuous real-valued function on a countably compact space can attain its maximum and minimum. However, the proof is not given. I cannot make a proof, and have done some search but failed to find any related consequence.
So I wonder is the proposition true? Anyone can provide a proof?
Let $X$ be countably compact, and let $f:X\to\Bbb R$ be continuous. Suppose that $f$ is bounded, and let $u=\sup\{f(x):x\in X\}$. Suppose that $u$ is not in the range of $f$, so that $f$ does not achieve its maximum. For each $n\in\Bbb Z^+$ let
$$U_n=\left\{x\in X:f(x)<u-\frac1n\right\}\;;$$
then $\mathscr{U}=\{U_n:n\in\Bbb Z^+\}$ is a countable open cover of $X$, so it has a finite subcover $\mathscr{U}_0$. Let $m=\max\{n\in\Bbb Z^+:U_n\in\mathscr{U}_0\}$; $U_n\subseteq U_{n+1}$ for each $n\in\Bbb Z^+$, so $X=\bigcup\mathscr{U}_0=U_m$, and therefore $f(x)<u-\frac1m$ for every $x\in X$, contradicting the choice of $u$. Thus, if $f$ is bounded, it must attain a maximum value. Applying this result to $-f$, we see that a bounded function must also attain its minimum value. Thus, we’re done if we can show that $f$ must be bounded.
This can be done with the same kind of argument. For each $n\in\Bbb Z^+$ let $U_n=\{x\in X:f(x)<n\}$, and let $\mathscr{U}=\{U_n:n\in\Bbb Z^+\}$; then $\mathscr{U}$ is a countable open cover of $X$, and $U_n\subseteq U_{n+1}$ for each $n\in\Bbb Z^+$, so there is an $m\in\Bbb Z^+$ such that $X=U_m$. But then $f(x)<m$ for each $x\in X$, and $f$ is bounded above. Since $-f$ is also bounded above, $f$ is bounded below and therefore bounded.
