Prove the cross of two open sets is also an open set.

Question

$U$ and $V$ are open subsets of $\mathbb{R}$. Show that $U×V = \{(x,y)\mid x\in U, y\in V \}$ is an open subset of $\mathbb{R}^{2}$.

I know the definition of an open subset is that $\forall (x,y) \in U×V, \exists \delta > 0$ such that $B\left((x,y), \delta\right) \subset U×V$.

I just don't know how to use that in this case.

Answer 1

Let $(x,y)\in U\times V$ be given.

Note that $x\in U$ and $y\in V$, so there exists $\delta_x >0$ and $\delta_y>0$ such that $$B(x,\delta_x)\subseteq U \iff \text{ and } B(y,\delta_y)\subseteq V.$$

Note that whenever $a\in B(x,\delta_x)$, we have $|x-a|<\delta_x$. And similarly $|y-b|<\delta_y$ whenever $b\in B(y,\delta_y)$.

Let $\delta =\min(\delta_x,\delta_y)$.

Now let $(a,b)\in B((x,y),\delta)$, then we have $\sqrt{(x-a)^2+(y-b)^2}<\delta$.

It implies that $|x-a|<\delta\leq \delta_x$ and that $|y-b|< \delta \leq \delta_y$ which means that $a\in U$ and $b\in V$, so $(a,b)\in U\times V$.

Therefore, $B((x,y),\delta)\subseteq U\times V$ and we are done.