Prove: If $A \times B$ is open in $X \times Y$ then $A$ is open in $X$ and $B$ is open in $Y$

Question

Suppose that $A \subset X$ and $B \subset Y$ where $X$ and $Y$ are nonempty topological spaces, and assume that $A \times B$ is a nonempty open subset of $X \times Y$.

Prove that $A$ is open in $X$ and $B$ is open in $Y$.

(Hint: Take intersection with $\{x_0\} \times Y$ and $X \times \{y_0\}$ where $(x_0, y_0) \in A \times B$ and the vertical and horizontal slices are homeorphic to the factors $X$ and $Y$.)

This is where I'm stuck. How do I use the assumption that $A \times B$ is an open subset of $X \times Y$ and use the hint to prove that $A$ and $B$ are open in $X$ and $Y$ respectively?

Answer 1

You could use that the projection map $\pi:X \times Y \to X$ is an open map. That means that the image of an open set via that projection is an open set of $X$. In particular, $A$ is an open set of $X$. The same can be done for $B$.

Answer 2

A base for the product topology on $X \times Y$ is given by $\{O_1 \times O_2: O_1 \in \mathcal{T}_X, O_2 \in \mathcal{T}_Y\}$.

Let $A \times B$ be open and non-empty. Pick any $(x_0,y_0) \in A \times B$. Proof that $A$ is open ($B$ is similar):

Pick $p \in A$. Then $(p, y_0) \in A \times B$ and as this set is open, we can find a basic open set $O_1 \times O_2$, with $O_1$ open in $X$ and $O_2$ open in $Y$, such that

$$(p,y_0) \in O_1 \times O_2 \subseteq A \times B$$

So $p \in O_1$ and $O_1 \subseteq A$ (as $p' \in O_1 \implies (p',y_0) \in O_1 \times O_2 \subseteq A \times B \implies p' \in A$) so $p$ is an interior point of $A$. As $p \in A$ was arbitrary, $A$ is open.

$B$ open goes the same way.