Exercise from book:
Let $\left(f_{k}\right)_{k=1}^{\infty}$ be a sequence of functions and suppose that they are all upper semi-continuous at $x_{0}$. Define the function $g$ by $g(x)=\inf _{1 \leq k<\infty} f_{k}(x)$. Show that $g$ is upper semi-continuous at $x_{0}$.
My attempt: By definition of upper semi continuity we have
$\lvert x-x_0\rvert \Longrightarrow f_{k}(x) < f_{k}(x_0)+\varepsilon$
$ \inf f_{k}(x) < f_{k}(x_0)+\varepsilon$
I think it completes the proof. Is it right or I missing something?????
I am not familiar with topology. so I am interested in only real analysis terms
Let $\varepsilon > 0$ and $g(x) = \inf_{k \in \mathbb N} f_k(x)$. Therefore, there exists $k_0 \in \mathbb N$ with $$f_{k_0}(x_0) - g(x_0) < \frac{\varepsilon}{2}.$$ Moreover, for such a $k_0$, there is $\delta_{k_0}> 0$ with the property that $$|x - x_0| < \delta_{k_{0}} \quad \Rightarrow \quad f_{k_0}(x) < f_{k_0}(x_0) + \frac{\varepsilon}{2}.$$ We deduce that, if $|x - x_0| < \delta_{k_0}$, $$g(x) \le f_{k_0}(x) < f_{k_0}(x_0) + \frac{\varepsilon}{2} < g(x_0) + \varepsilon.$$