Suppose fi is continuous from a topological space X to the real numbers,how to prove that g=sup{fi:i=1...n} is also continuous?
It is quite clear to show that a function is continuous we must only show that it's continuous with respect to some open subbase for the codomain. So, in particular since if $\displaystyle K=\left\{(a,\infty):a\in\mathbb{R}\right\}$ and $\displaystyle L=\left\{(-\infty,b):b\in\mathbb{R}\right\}$ then $\displaystyle K\cup L$ forms an open subbase for $\displaystyle \mathbb{R}$ we must only check the cases for elements of this set.
If $\displaystyle E\in K$ then $\displaystyle E=\left(a,\infty\right)$ and so $\displaystyle g^{-1}\left(E\right)=\left\{x\in X:\max_{1\leqslant j\leqslant n}f_j(x)>a\right\}=\bigcup_{j=1}^{n}\left\{x\in X:f_j(x)>a\right\}$ and since this is a union of open sets (since each $\displaystyle f_j$ is continuous) it follows that $\displaystyle g^{-1}\left(E\right)$ is continuous. The case for $\displaystyle E\in L$ is analogous.