On $C[0,1]$, consideer the norm $\|f\|= \max \{|f(t)| : t \in [0,1] > \}$. Let the continuous function $k : [0,1] \to (0, \infty)$ and let $$A = \{f \in C[0,1]: |f(t)| < k(t), \forall t \in [0,1]\}.$$ Show that $A$ is open.
I know that we look for $r > 0$ such that $B(f_0,r) \subset A$, but I really don't know how to start well the problem. I know that as $[0,1]$ is a compact set in $\mathbb{R}$, we can $$\|f - f_0\| < r \implies \max_{t \in [0,1]} |f_0(t)| < \max_{t \in [0,1]} |f(t)| + r.$$
Is anyone could help me at this point?
Suppose $f \in A$. Look at the function $\phi(t)= k(t)-|f(t)|$. Note that $\min_t \phi(t) >0$. How does this help?
Let $\epsilon = {1 \over 2} \min_t \phi(t)$ (which is a positive number, since $k-|f|$ is continuous). Then show that if $g \in B(f,\epsilon)$ that $g \in A$.