Definition of lower semi-continuous: Give topo space X and mapping $f:X \to \left(-\infty,+\infty\right]$. $f$ is lower semi-continuous at $x_0$ if $\forall \varepsilon > f(x_0)$, $\exists V$ is neighborhood of $x_0$ that if $\forall x\in V$ then $\varepsilon<f(x)$.
Problem: If $f$ is semi-continuous on compact space $X$ then $\exists x_0\in X$ that $\displaystyle f(x_0)=\min_{x\in X}f(x)$.
My attempt:
Set $a=\inf{f(X)}$. Choose decreasing sequence $\left\{a_n\right\}$ that $\lim\limits_{n\to \infty}a_n=a$, set $A_n=\left\{x\in X: f(x)\le a_n\right\}$. So we have if $\displaystyle x_0 \in \bigcap_{n=1}^{\infty}A_n$ then $f(x_0)=a$, this yields $\displaystyle f(x_0)=\min_{x\in X}f(x)$.
I wondering my proof is correct or not? Because I still not use "$X$ is compact" and I still not prove $f(X)$ is lower bounded.
Thanks for helping me.
Set $a=\inf{f(X)}$. Choose decreasing sequence $\left\{a_n\right\}$ that $\lim\limits_{n\to \infty}a_n=a$, set $A_n=\left\{x\in X: f(x)\le a_n\right\}$. We can see that $A_{n} \supset A_{n+1}$ for all $n\in \mathbb{N}$, this yields $\displaystyle \bigcap_{j \in J}A_j \neq \varnothing$ for all $J$ is finite in $\mathbb{N}$. Since $X$ is compact space and $A_n$ is closed in X for all $n\in \mathbb{N}$ then we have $\displaystyle \bigcap_{n=1}^{\infty}A_n \neq \varnothing$.
So we have if $\displaystyle x_0 \in \bigcap_{n=1}^{\infty}A_n$ then $f(x_0)=a$, this yields $\displaystyle f(x_0)=\min_{x\in X}f(x)$.