Is this correct $x\in\operatorname{supp}(f.g)\implies (f.g)(x)≠0\implies f(x)≠0 \operatorname{and} g(x)≠0$

Question

I'm going to understand those implications is correct or no

Let :

$f,g$ are two function then :

$$x\in\operatorname{supp}(f.g)\implies (f.g)(x)≠0\implies f(x)≠0 \operatorname{and} g(x)≠0$$

For example take :

$$f(x)=\begin{cases}1,x\in[0,1]\\0,\operatorname{otherwise}\end{cases}$$

$$g(x)=\begin{cases}1,x\in[2,3]\\0,\operatorname{otherwise}\end{cases}$$

Then :

$$\operatorname{supp}f=[0,1]$$

$$\operatorname{supp}g=[2,3]$$

So not always : $f(x)≠0$ and $g(x)≠0$

I already to see your exploration !

Answer 1

Since $\operatorname{supp}(h)$ is defined as the closure of $\{x\mid h(x) \neq 0\}$, the implication $x \in \operatorname{supp}(h) \implies h(x) \neq 0$ might be false on the boundary of $\operatorname{supp}(h)$.