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Given $X\subseteq \Bbb R^m, f:X\to\Bbb R$ and $x\in X$, we say $f$ is lower semicontinous (l.s.c for short) at x if
$\forall \varepsilon>0\ \exists\ \delta >0\ \forall \in B(\delta,x), \ f(x)\le f(y)+\varepsilon$.
If x is a limit point of $X$, then f is lower semicontinuous iff $f(x)\le \lim_{\delta\to0^+}\ \inf\ f{\restriction}(X\cap B^*(\delta,y))$. And, it is also given that $A\subseteq\Bbb R^m\times\Bbb R$, put $\pi(A)=\{x\in \Bbb R^m:\exists p\in \Bbb R,(x,p)\in A\}$, the projection of $A$ on $\Bbb R^m$. For $x\in\Bbb R^m$, put $A_x=\{p\in \Bbb R:(x,p)\in A\}$, the fiber of $A$ over $x$.

Then, I need to show:
if $A\subseteq\Bbb R^m\times\Bbb R$ is compact and $g:\pi(A)\to\Bbb R$ is defined by $g(x)=\inf\ A_x$, then g is lower semicontinuous.

I got some idea about this problem, which is bu using the iff statement of limit inferior to show if g is not lsc, then it contradicts to the condition that $A$ is not compact. But I just do not know how to show this formally.

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1 Answer 1

Assume that the iff statement is violated. Then you get $\varepsilon > 0$ and a sequence $x_n \to x$ with $g(x) > g(x_n) + \varepsilon$.. Now this should help you to get a contradiction with the definition of $g$ / the compacity of $A$.

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