Define $g=\inf_sg_s$. If $g_s$ is upper semicontinuous then so is $g$?

Question

Define $g_s(x)=\begin{cases}1\quad x\in\overline {V_s}\\s\quad o.c. \end{cases}$

$s\in(0,1).$

$g_s$ is upper semicontinuous because $\{x:g_s(x)<a\}$ for all $a$ real is open.

Can we still say that $g=\inf_sg_s$ is upper semicontinuos?

Answer 1

$\{x:\inf g_s (x) <a\}=\cup_s \{x:g_s(x) <a\}$. since union of open sets is open it follows that $\inf_s g_s$ is upper semicontinuous.

Answer 2

Yes, the claim is true: fix $a\in\mathbb R$ and let $x_0\in V=\{x:g(x)<a\}$. By hypothesis, there is a $g_s$ such that $g_s(x_0)<r<a.$ And since $g_s$ is upper semicontinuous, there is an open set $U\ni x_0$ such that $x\in U\Rightarrow g_s(x)<r$. Thus, if $x\in U$, then $g(x)<g_s(x)<r<a$ so $V$ is open and so $g$ is upper semicontinuous.