Open subset of C[0,1]?

Question

Let us define $\|u\|_{\infty}=\sup_{x\in[0,1]}|u(x)|$ in the space $C[0,1]$. So we are working in the normed vector space $(C[0,1], \|\cdot\|_{\infty})$. $\;$Let: $$\boldsymbol{F}=\{f\in C[0,1]: f(x)>0,\; \forall\; x\in [0,1]\}.$$ Is $\boldsymbol{F}$ open?

Intuition says yes it is open because of the strict inequality, but then I thought that can't I just take a function $f\in \boldsymbol{F}$ with $\|f\|_{\infty}=a$ and then take a function $g\notin \boldsymbol{F}$ but $\|g\|_{\infty}=a-\varepsilon$ so the difference is arbitarily small and this would say that $\boldsymbol{F}$ is not open?

Are my ideas correct? And if so can someone maybe give more insight on how formulate this proof nicer and also how to think about this space geometrically.

Answer 1

Let $f\in F$; since $f$ is continuous and takes only positive values, the infimum of $f$ on $[0,1]$ is also positive, say equal to $m$. The ball for the uniform norm centered at $f$ and having radius $m/2$ is contained in $F$, hence $F$ is open.

Answer 2

To show that it's open we need to show that around ever $f \in F$ there is a ball $B \subset F$ containing $f$. The trick here is not only the strict inequality, but the compactness of the interval $[0, 1]$. If $f \in F$ then $\min(f) > 0$. Then take the ball centered at $f$ of radius $\min(f)/2$.