Show $A = \{f \in C([0,1]): f(x) > 0 \forall x \in [0,1] \}$ is open w.r.t sup norm: $\|f\|_{\infty, [0,1]} = sup_{x \in[0,1]}|f(x)|$ Pf Verification

Question

Show $A = \{f \in C([0,1]): f(x) > 0 \forall x \in [0,1] \}$ is open w.r.t sup norm: $\|f\|_{\infty, [0,1]} = sup_{x \in[0,1]}|f(x)|$

I wanted to know if I showed it properly.

Attempt:

So what we want to show is that $\forall \ g(x) \in A$ there exists $\delta > 0$ s.t $B_{\delta}(g) = \{h: \|g-h\|_{\infty} < \delta \} \subset A$.

By the Extreme Value Theorem all $f \in A$ attain a max and min, also we are given $f(x) > 0$ this means there exists $\delta > 0$ s.t $g(x) > \delta > 0 \ \forall x \in [0,1]$. Equivalently: $g(x) - \delta > 0$

So in order to satisfy $B_{\delta}(g) \subset A$, I need to show for all functions $h(x) \in B_{\delta}(g)$ that $h(x) > 0$.

Here is where I'm not sure if I am doing the right thing.

Consider $\|h-g\|_{\infty} < \delta$

$\Rightarrow sup_{x \in [0.1]}|h(x) - g(x)| < \delta$

$\Rightarrow -\delta < h(x) - g(x) < \delta \\ \Rightarrow h(x) > g(x) - \delta > 0$

This shows that $h(x) \in B_{\delta}(g) \subset A$

So the set $A$ is open.

Remark: This solution that I wrote doesn't "feel" complete to me> I used all of the conditions of the set, but I' feeling I missed something in my logic.

Answer 1

I'm not sure that there's much more to say to this other than that your proof is correct.