Showing that $M := \{f \in C([-1,1]) : f(0) > 0\}$ is open

Question

Given $f \in C([-1,1],\mathbb{R})$ equipped with sup norm metric. I am trying to find out whether the subset $$M := \{f \in C([-1,1]) : f(0) > 0\} $$ is open/closed in $C([-1,1])$

My work:

$M$ is open. Since any function in $M$ doesn't cross the negative y-axis, we can find an $\epsilon$-ball around the axis, such that $B_{\epsilon}(f)$ is completely contained within $M$. I am having trouble formalizing this, i.e finding such $\epsilon$

$M$ is not closed. Consider the sequence of functions $f_{n}(x) := \frac{x+1}{n}$

$f_{n}(0) > 0$ for all $n \in \mathbb{N}$ but $f_{n}$ converges to the zero function, which is not an element of $M$. So $M$ doesn't include its limit points.

Answer 1

The set is indeed open. If $f \in M$, then $f(0) = \epsilon > 0$. But then if $\|g -f \|_\infty < \frac{\epsilon}{2}$ , we will find that $g(0) \ge \frac{\epsilon}{2} > 0$. This shows that $M$ contains a ball around of $f$ of radius $\epsilon/2$, and so $M$ is open.

Answer 2

The map $\phi: C([-1,1]) \to \mathbb R$ given by $\phi(f) = f(0)$ is continuous, since $|f(0)-g(0)| \le \|f - g\|$. Since $(0,\infty)$ is open in $\mathbb R$, $M = \phi^{-1}((0,\infty))$ is open in $C([-1,1])$.