Relation between a set and its interior Let $C[0,1]$ be a space of continuous functions $f: [0,1] \rightarrow \mathbb R$ with metric $d(f,g)=\sup |f(x)-g(x)|$. Find the interior of
$$A= \{ f \in C[0,1] : \exists t\in [0,\frac{1}{2} ] \ f(t)=0 \}$$
$$B= \{ f \in C[0,1] : \exists t\in [0,\frac{1}{2}) \ f(t)=0 \}$$
My solution:
$\operatorname{Int}(B)=B$, because for every function $f \in \operatorname{Int}(B)$ I can can set $r$ so small that the distance between that functions and functions in $B(f,r)$ have to fulfill the condition $\exists t\in [0,\frac{1}{2}) \ f(t)=0$. Is this correct?

$$\mathrm{Int}(A)=\{f\in C[0,1]\mid\exists (x,y)\in[0,1/2]^2,f(x)f(y)\lt0\}=\mathrm{Int}(B)\quad(\ne\varnothing)$$
