Vtyjkyuk

Show that the supremum of any collection of lower semicontinuous functions is lower semicontinuous.

Suppose $f_n$ is a sequence of lower semicontinuous functions on a topological space $X$. Define $$g_k=sup_nge kf_n.$$ I could see that $$x: g_k(x)gt alpha=bigcup_n=k^inftyx:f_n(x)gt alpha$$ from where it follows that $g_k$ is lower semicontinuous. Also $g_n$ is monotonically decreasing sequence of semicontinuous functions, and because $$limsup_nrightarrowinfty f_n(x)=inf g_1(x), g_2(x), cdot cdot cdot$$, $limsup_nrightarrowinfty f_n(x)$ is lower semicontinuous if it can be shown that a monotonically decreasing sequence of lower semicontinuous functions converges to a lower semicontinuous function. How can I show this?

Also, even if the above can be implemented, I would only prove that the supremum of a sequence of lower semicontinuous functions is lower semicontinuous. What's the way generalize this to "any collection" of lower semicontinuous functions?

Thanks.

Note that a function $f$ is lower semicontinuous iff given $x in X$ and $r < f(x)$ there is an open neighborhood of $U$ of $x$ such that $r < f(y)$ for each $y in U$.

Suppose that $ f_i _i in I$ is any collection of lower semicontinuous functions on $X$, and let $g$ be the pointwise supremum, i.e., $g (x) = sup_i in I f_i(x)$ for all $x in X$.

Taking $x in X$, and any $r < g(x) = sup_i in I f_i(x)$ it must be that there is an $i in I$ such that $r < f_i(x)$. Since $f_i$ is lower semicontinuous there is an open neighborhood $U$ of $x$ such that $r < f_i(y)$ for each $y in U$. As $f_i(y) leq g(y)$ for all $y$, it follows that $r < g(y)$ for each $y in U$.