about $C([0,1])$ with sup metric Define the space $C([0,1])$ as the space of continuous functions $f : [0,1] \mapsto \Bbb R$ with   $C([0,1])$ $$ d(f,g) = \sup _{x \in [0,1]}{|f(x)-g(x)|} ,  $$
so  let $$A= \left\{f \in C([0,1])\ \middle| \ 0 <\int_0^1 f(x)  \ \mathrm{d}x  < 1\right\}$$  now is $A$ open , close , bounded , connect or compact ? 
I think $A$ is open because for every $f \in A $ we have $B_t (f) \subseteq A $ such that $t:= 1- \int_0^1 f(x)  \ \mathrm{d}x $.(note that $B_t (f) $ is open ball with center $f$ and radios $t)$. $A$ is not close because if we let $f_n (x)= \frac{1}{n}$ then $ 0< \int_0^1 f_n(x)  \ \mathrm{d}x=\frac{1}{n} <1 $ and for every $1< n \in \mathbb{N}$ ,$  f_n(x) \in A$ and $lim_{n \to \infty} f_n(x)=0$  then $  \int_0^1 lim_{n \to \infty} f_n(x)  \ \mathrm{d}x=0  $ then 
 $ lim_{n \to \infty} f_n(x) \notin A$ hence $A$ is not close and yet $A$ is not compact .

$A$ is not bounded
The piecewise linear map defined by $f_n(0)=f_n(1/n)=f_n(1)=0$ and $f_n(1/(2n))=n$ is such that $\int_0^1 f_n = 1/2$ but $d(f_n,0) = n$ is unbounded.
$A$ is connected
$A$ is connected because it is convex.
