Somemathhere

Let $X,Y$ be topological spaces, $Y$ is compact and $f:X\times Y\to \mathbb{R}$ be a continuous function. Define $$g(x)=\inf_{y\in Y}f(x,y)$$ Then $g$ is continuous. $\textbf{Proof}$. It is enough to check that for all real $a$ the sets $g^{-1}((-\infty,a))$ and $g^{-1}((a,\infty))$ are both open in $X$. First rewrite $$g^{-1}((-\infty,a))=\{x\in X\ | \ \inf_{y\in Y}f(x,y)<a\}=\{x\in X\ | \ \exists y\in Y, f(x,y)<a\}.$$ Observe that $$\{x\in X\ | \ \exists y\in Y, f(x,y)<a\}=\bigcup_{y_0\in Y}\{x\in X\ | \ f(x,y_0)<a\}.$$ Since $f$ is continuous, the function $h_{y_0}(x)=f(x,y_0)$ is continuous as well. This means that the set $\{x\in X\ | \ f(x,y_0)<a\}$ is open (being preimage of the open interval $(-\infty,a)$ under the continuous function $h_{y_0}$). Hence $\{x\in X\ | \ \exists y\in Y, f(x,y)<a\}$ is a union of open sets - thus it is open. Now $$g^{-1}((a,\infty))=\{x\in X\ | \ \inf_{y\in Y}f(x,y)>a\}.$$ Denote $$V_n=\left\{x\in X\ | \ \forall y\in Y,\...