I have a problem about the definition of lower semi-continuous (lsc) function and minimize problem:
Is a function defined on a compact set which is lsc will has FINITE lower bound?
In many books, this is a celebrated theorem. But consider $f(x)=-\frac{1}{x}$ on $[0,1]$. $f$ has extended value at $x=0$ that $f(0)=-\infty$. Obviously, $f(x_0)\leq \liminf\limits_{x\rightarrow x_0}f(x)$ for $\forall x_0\in[0,1]$. In fact, the '$\leq$' can be '$=$'. I confuse this because I see a proof on a book which use the other definition of lsc that:
$f:X\mapsto R$ is lsc iff $f^{-1}(a,\infty)$ is open in $X$ for $\forall x\in R$.
Then, to prove every lsc function defined on compact set has finite lower bound, they use $X=\bigcup_{n=1}^\infty f^{-1}(-n,\infty)$. However, in my example, this is wrong. Because $[0,1]=X\neq \bigcup_{n=1}^\infty f^{-1}(-n,\infty)=\bigcup_{n=1}^\infty (\frac{1}{n},1]=(0,1]$. I think I misunderstand some definitions here but I do not know where is my mistake. Could you help me to point out? Thank you very much in advance.
Your function does not map $[0,1]$ to $\mathbb{R}$, but to $\mathbb{R} \cup \{-\infty\}$. Hence, it is not lsc according to your definition, which requires $f \colon X \to \mathbb{R}$.