Let $(C[0,1],d_{\text{sup}})$ – space of continuous functions from $[0,1]$ to $\mathbb R$ with a "supremum metric": $$d_{\text{sup}}(f,g)=\text{sup} \left\{|f(t)-g(t)|:t\in [0,1] \right\}$$Find closure and interior for the set:
$$A=\left\{f\in C[0,1]:f(t)>0, t\in[0,1] \right\}$$ in the topology generated by the supremum metric.
My try:
Let $$F=\left\{f\in C[0,1]:f(t)\ge0, t\in[0,1] \right\}$$
We should prove that $\overline A=F$. So:
$$A \subset F \Rightarrow \overline A \subset \overline F=F \subset X \setminus F = \left\{ f \in C[0,1]: \exists t\in [0,1]: f(t)<0 \right\}$$
However I don't know what I can do the next. Moreover I have a problem with a interior.
Best Answer
Your set $A$ is open. In fact, if $f\in A$ and $m=\min f$, then $B_m(f)\subset A$. So, $\mathring A=A$.
You are right: $\overline A=F$. The set $F$ is closed, because if $f\notin F$, then there is some $t\in[0,1]$ such that $f(t)<0$. But then if $g\in B_{|f(t)|/2}(f)$, you have $g(t)<0$ and therefore $g\in F^\complement$. So, $F^\complement$ is open, which means that $F$ is closed. So, since $A\subset F$, $\overline A\subset F$.
And is $f\in F$ and $r>0$, consider the function $g=\max\{f,r\}$. It belongs to $C[0,1]$. Actually, $f\in A$ and $d_{\text{sup}}(f,g)\leqslant r$. So, $f\in\overline A$. This proves that $F\subset\overline A$. Since $\overline A\subset F$, this proves that $\overline A=F$.