I have to prove that the set $A=\{ f \in C[0,1] |\quad ||f|| >1 \}$ is open in $C[0,1]$. The norm is standard supremum norm. I tried this:
Let $f \in A$. Then $ \sup_{ t \in [0,1]} |f(t)| > 1$. Since $f$ is continuous on compact set, then there is some $t_0 \in [0,1]$ such that $|f(t_0)|= \sup_{ t \in [0,1]} |f(t)| > 1$. Also, there exists $\epsilon>0$ such that $|f(t_0)| > 1+ \epsilon$. Let $g \in B(f, \epsilon)$. Then $||f - g||= \sup_{ t \in [0,1]} |f(t) - g(t)| < \epsilon$.
We have to prove that $||g|| > 1$. Then we have:
$$||g||= || g- f + f|| \geq ||f|| - || g-f|| = |f(t_0)| - || g-f|| > 1 + \epsilon - \epsilon = 1$$
Hence, $ g \in A $.
Is this method correct? Thanks in advance!
$$A=||\cdot ||^{-1}((1, \infty))$$
But the norm function $||\cdot ||\colon C([0,1])\to \mathbb{R}$ is continuos with respect the topology induced by its distance, thus $A$ is open.