$\{f\in C[a,b]:f(x)>0\}$ is open

Question

Show that the set $\{f\in C[a,b]:f(x)>0$ for all $x\in[a,b]\}$ is open in $C[a,b]$.

Let me first try by talking about open sets definition. The set of functions greater than $0$ must admit near functions that are also greater than $0$ for some fixed funciton $f$. However, I don't have a norm to induce a metric and measure things here.

So I need to try other way. I think it may have something to do with this set being the inverse of an open set by some continuous function, but I cannot think exactly which inverse it would be.

Answer 1

Let $f_0$ be an element in the given set $X$ of all functions $f$ in $C[a,b]$ with $f>0$.

(We will find an $\epsilon>0$, so that the ball $B(f_0,\epsilon)$ around $f_0$ of radius $\epsilon$ with respect to the norm $\|\ \cdot\ \|_\infty$ lies fully inside $X$.)

Let $m=\min f>0$ be the minimal value taken by $f$ on $[a,b]$.

Set $\epsilon =m/2>0$. Then the ball around $f_0$ with this radius is fully included in $X$.

So $X$ is open (by definition). $\square$

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Alternatively, we can write: $$ X=\bigcup_{f>0} B(f,\min f/2) $$ as a union of open balls.

Answer 2

It's easy to verify that a set $O$ is open if and only if for every sequence that converges to an $x \in O$ there exists $N \in \mathbb{N}$ such that for $n \geq N$, we have $x_n \in O$. Let $f_n \rightarrow f$ be such a sequence. Since $f$ is continuous and $[a,b]$ is a compact set, there exist $y, Y > 0$ such that for all $x \in [a,b]$, we have $y \leq f(x) \leq Y$. Now, recall that $(f_n)$ converges uniformly and use the uniform convergence of $(f_n)$ to show that there exists an $N$ that satisfies our condition.