Can $f \in C[0, 1]$ with a countable number of zeros be a zero divisor? Let us consider the ring of all continuous functions $ C[0,1]$. Let $f \in C[0,1]$ be such that $f$ has countable number of zeros.

Can $f$ be a zero divisor?

Surely $f$ can't be an unit. If it is a zero divisor then there must exist an element $g \neq 0$ such that $fg \equiv 0$. That is, $g$ must vanish on uncountable number of points in $[0,1]$. But I can't figure out if it's true.
 A: Suppose that $f$ is a zero divisor, then there is $g \in C[0, 1]$, $g \neq 0$, such that $fg = 0$. That is, for every $x \in [0, 1]$, $(fg)(x) = f(x)g(x) = 0$. So, for every $x \in [0, 1]$ with $f(x) \neq 0$, we must have $g(x) = 0$; in particular, $g$ takes non-zero values on at most countably many points. 
Denote the set of zeroes of $f$ by $A$. As $g$ must vanish on the complement of $A$, but $g \neq 0$, there is $a \in A$ such that $g(a) \neq 0$; without loss of generality, let's suppose $g(a) > 0$. As $g$ is continuous at $a$, there is $\delta > 0$ such that $g(x) > 0$ for $x \in (a - \delta, a + \delta)$ which contradicts the fact that $g$ is non-zero on at most countably many points.
A: Here $f$ cannot be a zero divisor. Suppose that $fg$ is zero and $g$ is nonzero. Clearly $g(x) = 0$ whenever $f(x) \neq 0$.
Let $c \in [0,1]$ be a zero of $f$. Suppose that $g(c) \neq 0$. By continuity of $g$, produce an interval $(c - \delta, c + \delta)$ on which $g$ is nonzero. Then $f$ must be zero on $(c - \delta, c + \delta)$. This contradicts that $f$ has countably many zeroes.
