The set of continuous functions positive at $0$ dense in $C[0,1].$

Question

I read a comment in a text on functional analysis that the set $\{f\in C[0,1]:f(0)>0\}$ is a dense subset of $C[0,1]$ with respect to the norm $\|f\|:=\max\limits_{0\leqslant x\leqslant1}|f(x)|.$

I can't see why this should be true. For instance, couldn't I just take function $g\in C[0,1]$ such that $g(0)<0,$ making it impossible for a sequence $(f_n)_{n=1}^\infty$ such that $f_n(0)>0$ for all $n$ to tend to $g$? I don't see how any neighbourhood of $g$ could contain almost all $f_n$.

If someone could help explaining why this should be true I would be grateful. Thanks.

Answer 1

You are correct, the statement as it stands is wrong, for the reason you are suggesting.

Answer 2

You are absolutely right, a way to see it is that: $$\phi:(C([0,1]),\|\|) \to \Bbb R), f \mapsto f(0)$$ is a continuous linear form so to rephrase your agument, $$\phi^{-1}(\{x >0\}) \cap\phi^{-1}(\{x <0\})=\emptyset$$ but the second one is open and non empty so the first one cannot be dense.