I've proved the following statement: "Given $(Z, T_Z)$ the product topological space of $(X, T_X)$ and $(Y, T_Y)$. If $A \in T_X$ and $B \in T_Y$ then $A \times B \in T_Z$".
Converse statement: "Given $(Z, T_Z)$ the product topological space of $(X, T_X)$ and $(Y, T_Y)$. If $A \times B \in T_Z$ then $A \in T_X$ and $B \in T_Y$".
Now, I'm trying to prove the converse of the statement is not always true but I cannot take find an example to show it's false.
Is the converse of the statement true? If not, please find an example to show it's false.
P/S: I'm thinking the statement is not true because of the union operation of cartesian.
If $A$ is empty and $B$ is not open it follows that $A\times B$ is open in the product space. So the converse is false. But if $A$ and $B$ are nonempty and $A \times B$ is open then $A$ and $B$ are open. Proof: Define $f: X \to Z$ by $f(x)=(x,b)$ where $b$ is fixed element of $B$. Show that this map is continuous. The inverse image under this map of $A\times B$ is exactly $A$ so $A$ is open. Similarly, $B$ is also open.