Showing that $B=\{(x,y) \in \Bbb{R^2} \mid \ xy>0 \}$ is an open set 
Show that $$B=\{(x,y) \in \Bbb{R^2} \mid \ xy>0 \}$$ is an open set in the Euclidean space $(\Bbb{R^2},d_2),$ where $d_2(x,y)=\sqrt{(x_1-y_1)^2+(x_2-y_2)^2}$.

Defining $A= \{(x,y) \in \Bbb{R^2} \mid x>0,y>0 \}$ and $E=\{  (x,y) \in \Bbb{R^2} \mid x<0,y<0 \}$ one sees that $B= A\cup E$ and thus $B$ will be open if $A$ and $E$ are open.
How can I show that $A$ and $E$ are both open? It seems that I would need to find open balls with arbitrary radius $B((x,y), r)$ for both cases that are contained in the set I'm working with?

You can show $B^C$ is closed.
In fact $B^C = \{x,y\in \mathbb{R}^2,xy\leqslant 0\}$.
If you take $(x_n,y_n)$ a sequence of elements of $B^C$ that converges to $(x,y)$, then for all $n$, $x_ny_n \leqslant 0$.
When $n\rightarrow +\infty$, you get $xy \leqslant 0$.
So $(x,y)\in B^C$.
So $B^C$ is closed.
So $B$ is an open set.
