# Intersection of open and lower semicontinuous correspondence is lower semicontinuous

I have a little problem in proving this. So, suppose we have two multifunctions $f, g: E \to F$, where f is lower semicontinuous and g has open graph. I need to show that intersection of those (of images precisely) is lower semicontinuous.

The proof goes this way: We suppose that the intersection is non-empty. So let U be the open set that:

$$y \in (f \cap g)(x) \cap U.$$ We know that g has an open graph, so there exists neighborhood $$U_x \times U_y$$ of point (x,y) which lies in graph of g. In other words, the set $$\{(x,y) \in E \times F : y \in g(x) \}$$ is open in $E \times F$. We also know that f is lower semicontinuous, so $f^{-}[U] \cap U_x$ is an open neighborhood of x, where $f^{-}[U] = \left\{ x \in E: f(x) \cap U \neq \emptyset \right\}$. What now is enough to show is that $$f^{-}[U] \cap U_x \subset (f \cap g)^{-}[U].$$ So taking x from left side, we need to check if it satisfies that $f(x) \cap g(x) \cap U \neq \emptyset.$

And here I have a little problem, because when taking $\bar{x}$ which lies in the left side of inclusion, I know that f has non-empty intersection with U (via lower semicontinuity of f). Do you have any tips, how to show that the inclusion holds? For example, using the openness of graph of g? Any tips will be appreciated.

• Try to show that $f^{-}[U\cap U_y]\cap U_x\subset (f\cap g)^{-}[U]$. – Peter Elias Jun 20 '17 at 20:38