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?
 A: 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.
A: For $(x,y) \in A$ notice that
$$(x,y) \in B\left((x,y),\frac12\min\{|x|,|y|\}\right) \subseteq A$$
and the same thing for $E$.
For example, if $(x',y') \in B\left((x,y),\frac12\min\{|x|,|y|\}\right)$ we have
$$x-x' \le |x-x'| \le \sqrt{|x-x'|^2 + |y-y'|^2} \le \|(x,y)-(x',y')\| < \frac{x}2 $$
so $x'> \frac{x}2>0$. You can show the other inequalities similarly.
A: Note that $$B = (-\infty,0)\times(-\infty,0)\cup(0,\infty)\times(0,\infty)$$
Since each of these is an open set, so $B$ is an open set.
A: The multiplication $f\colon
\begin{cases}\mathbb R^2&\to\mathbb R;\\
(x,y)&\mapsto x\cdot y
\end{cases}$ is known to be continuous.  Observe that $B$ is the preimage of the open set $(0,\infty)$ under $f$.
A: Suppose $x>0, y>0,$ and suppose $\min\{x,y\}=y.$ [A very similar proof can be done if $\min\{x,y\}=x$ ].
I want to do better than mechanodroid and show that $B\left(\ (x,y),\ y\ \right) \subseteq A$ which amounts to:
If $\ d\left(\ (x,y),\ (x',y')\ \right) < y,\ $ then $x' > 0$ and $y' > 0.$
Since showing that $x'>0\ $ is easier than showing $y' > 0,$ I'll concentrate on $y'$ and leave $x'$ as an exercise for the reader.
If $y' \geq y,$ then we are done since $y>0.$ So suppose further that $y' < y.$
By supposition, \begin{align} 0 < \sqrt{(x-x')^2 + (y-y')^2} < y \\
\implies (x-x')^2 + (y-y')^2 < y^2 \\
\implies 0 \leq (x-x')^2 < 2yy' - (y')^2 = y'(2y-y').
\end{align}
If $y'=0$ then $(x-x')^2 < 0 \ \Rightarrow\Leftarrow.$
If $y' < 0,$ then we require $2y - y' < 0$, which is true $\iff 0< 2y<y'<y\ \Rightarrow\Leftarrow.$
Therefore $y'>0.$
