$f,g:[0,1]\to \mathbb R$ continuous function and $f(t)<g(t), \forall t\in [0,1]$
$$U:=\{h\in C[0,1]:\forall t \in [0,1]:f(t) <h(t) <g(t) \}$$
in the space $X:C[0,1],||||_{\infty}$
I wanted to show that U is open ?
I know that $r_1=inf|f(t)-h(t)|$
and $r_2=inf|g(t)-h(t)|$
If $r_1,r_2>0$ then I can take minima of that to show open but why $r_1,r_2>0$ ?
ANy help will be appreciated
Edit:
[0,1] is compact so $g(t)-f(t)$ is continuous and attain minima >0 so there minima is say r
Now similary $h(t)-f(t)$ is also continuous and attain minima which is also >0
If $h \in U$, then $h - f$ and $g - h$ are both continuous functions on $[0,1]$ with values in $(0,\infty)$.
So both assume minima by compactness of $[0,1]$, say $\min(h-f) = r_1 >0$ and $\min(g-h) = r_2 >0$.
It is then easy to check that for $r := \frac{1}{2}\min(r_1,r_2)$ we can see that if $\|h'-h\|_{\mathrm{sup}} < r$ we have that also $f(t) < h'(t) < g(t)$, so that the ball with radius $r$ around $h$ in the sup-metric lies entirely inside $U$, so that $h$ lies in the interior of $U$, and as $h \in U$ was arbitary, $U$ is open in topology induced by the sup-metric.