Is the Cartesian product of two open sets open? Just a quick question: If you have two sets $A,B \subset \mathbb{R}$ that are open, that is, for every $p \in A$, there exists an $\varepsilon > 0$ such that $B(p;\varepsilon) \subset A$, is the Cartesian Product of these sets also open?
I am trying to think of a proof, but I am stuck rather quickly. So far I have this:
Proof: Let $A$, $B \subset \mathbb{R}$ be open sets. Let $C = A \times B$. Since $A$ is open, there exists an $\varepsilon > 0$ such that for all $p \in A$, $B(p;\varepsilon ) \subset A$. Since $B$ is open, there exists an $\varepsilon ' > 0$ such that for all $q \in B$, $B(q;\varepsilon ') \subset B$.
In other words, we can make a ball around a point in the $x$ or $y$ direction, but can we also make a ball in both at the same time?
Some guidance would be lovely.
Thanks in advance!
 A: Yes, the product of two open sets is open. Generally, that is immediate from the definition of the product topology, but in the situation here, the task is to show that the topology induced by the usual Euclidean metric on $\mathbb{R}^d$ (here, $d = 2$) is at least as fine as the product topology (the converse, that every open ball around a point contains the product of open balls around the coordinates, then shows that the two topologies are identical).
If we have $B(p;\varepsilon) \subset A$ and $B(q;\varepsilon') \subset B$, then $B(p;\varepsilon)\times B(q;\varepsilon') \subset A\times B$. Since the coordinate projections don't increase distance,
$$\lvert x_i-y_i\rvert \leqslant \sqrt{\sum_{j=1}^d \lvert x_j - y_j\rvert^2} = \lVert x-y\rVert,$$
it follows that $B((p,q);\min \{\varepsilon,\,\varepsilon'\}) \subset B(p;\varepsilon)\times B(q;\varepsilon')$, and the product of the two open sets $A$ and $B$ is recognised as open.
The converse, that every open ball in the Euclidean metric contains the product of open balls in the coordinate spaces, follows from
$$\lvert x_i-y_i\rvert < \varepsilon/\sqrt{d} \text{ for all } i \Rightarrow \lVert x-y\rVert = \sqrt{\sum_{j=1}^d \lvert x_j-y_j\rvert^2} < \sqrt{\sum_{j=1}^d \varepsilon^2/d} = \varepsilon.$$
