Let max $\{f,g\}$ be the function defined by max $\{f,g\}(x)$ = max $\{f(x),g(x)\}$, for $x \in E$ show that max $\{f,g\}$ is continuous.
How do i use the definition of $\epsilon$, $\delta$ to show this provided that $f,g$ are both continuous function.
Hint:
$$\max\{f(x),g(x)\}=\frac{|f(x)-g(x)|}{2}+\frac{f(x)+g(x)}{2}$$
Hint: For the $\epsilon$-$\delta$ hassle, for all $\epsilon>0$ you get $\delta_f$ and $\delta_g$, and you simply choose $\delta_{\max\{f,g\}}\mathrel{:=}\min \{\delta_f,\delta_g\}$.
Of course, it may be simpler to prove that $x\mapsto |x|$ is continuous and then use the trick by Mario G, but the beauty is in the eye of beholder, right? :-)
Suppose you have continuous $f_k$, $k=1,...,p$. Let $\psi(x) = \max_k f_k(x)$.
Choose $x$ and let $\epsilon>0$, and find $\delta>0$ such that $-\epsilon < f_k(x)-f_k(y) < \epsilon$ for all $y \in B(x,\delta)$.
Rewrite as $f_k(x) < \epsilon + f_k(y) \le \epsilon +\psi(y)$, and now take the $\max$ of the left hand side to get $\psi(x) < \epsilon +\psi(y)$, or $\psi(x)-\psi(y) < \epsilon$.
Repeating for the other inequality gives $-\epsilon < \psi(x)-\psi(y) < \epsilon$ for all $y \in B(x,\delta)$, and so $\psi$ is continuous.
