Question:
$(x_n \rightarrow x \implies \limsup_{n \rightarrow \infty } f(x_n) \leq f(x) ) \implies \{x \in X |f(x) < c\} \,\text{is open for any}\, c \in R ?$
I know related questions have been touched open in upper semi-continuity functions , but most of them proved the opposite direction, and I haven't managed to find a satisfying proof for this one.
Can anyone help me with a simple but rigorous proof?
Related MSE Post:
Equivalence of definitions for upper semicontinuity
Thanks.
On
To show $\{x\in X : f(x) <c\}$ is open, it is enough to show that $B=\{x\in X : f(x) \ge c\}$ is closed.
Proof : Let $\alpha \in B'$ i.e $\alpha$ is a limit point of $B$ .
To show $\alpha\in B$ .
Since $\alpha\in B'$, $\exists (x_n) \in B$ such that $x_n\to \alpha$
Then by the hypothesis, $\limsup f(x_n)\le f(\alpha)$
As $(x_n) \in B$ , $f(x_n) \ge c$
$c\le \limsup f(x_n)$
Then $c\le \limsup f(x_n)\le f(\alpha)$
Hence $f(\alpha) \ge c$ implies $\alpha \in B$
Thus the set $B$ closed. (Proved!)
It can be easily proved by contradiction.
If $W_c=\{x\in X: f(x)<c\}$ is not open, for some $c\in\mathbb R$, then there exist a $\xi\in W_c$, i.e. $f(\xi)<c$, such that $B_\varepsilon(\xi)\not\subset W_c$, for every $\varepsilon>0$. In partiucular, for $\varepsilon=1/n$, there would exist an $\xi_n\in X$, such that $d(\xi_n,\xi)<1/n$, and $\xi_n\not\in W_c$, i.e. $f(\xi_n)\ge c$. But clearly, $\xi_n\to\xi$, and due to the given condition, we would have that $$ c\le \limsup f(\xi_n)\le f(\xi)<c. $$ Contradiction.