Open sets in $Z=$ {$(x, \phi(x)) \in X \times Y: x \in X $} Describe the open sets in $Z=$ {$(x, \phi(x)) \in X \times Y: x \in X $} where $\phi \in C(X,Y)$ and $X,Y$ are some topological spaces.
(not sure this is totally correct) Assume $D \subseteq Z$ an open set, then $D$ must have the form $D = (D_x,D_y)$ where $D_x \subseteq X$ is open and $D_y \subseteq Y$ is also open and I also know that $\phi(D) \subseteq Y$ is open iff $D \subseteq X$ is open. But still I'm not sure how to describe the open sets. 
Would be glad for any help!
 A: Hint: The map $(x,\phi (x)): X \to Z$ is a homeomorphism. Hence a set is open in $Z$ iff it is of the form $\{(x,\phi (x)): x \in U\}$ for some open set $U$ in $X$. 
Proof of the fact that $(x,\phi (x)): X \to Z$ is a homeomorphism: this map is easily seen to be bijective. To show that it is continuous it is enough to show that $\{x\in X: (x,\phi(x)) \in V_1\times V_2\}$ is open in $X$ whenever $V_1$ and $V_2$ are open in $X$ and $Y$ respectively. This is because sets of the form $V_1\times V_2$  form a base for the topology of the product space $X \times Y$). But this set is noting but $V_1 \cap \phi^{-1}(V_2)$ so it is open in $X$.  To show that it is a homeomorphism it is remains to show that the image of any open set is open. Let $U$ be open in $X$. The image of $U$ is $\{(x,\phi(x)): x\in U\}$. We can write this set as $Z\cap  (U\times Y)$. Since $U\times Y$ is open on $X\times Y$ it follows that $Z\cap  (U\times Y)$ is open in the subspace topology of $Z$. This completes the proof. 
