Prove that $\{(x,y)\mid xy>0\}$ is open I need to prove this using open balls. So the general idea is to construct a open ball around a point of the set. A point $(x,y)$ such that $xy>0$. Then we must prove that this ball is inside the set. However, I don't know how to find a radius for this open ball. Can somebody help me in this proof?
 A: Prove it using a different approach:
Let this set be named $A$. Then:
$$A^C = \{(x,y) \in \mathbb R^2, \ xy \le 0 \}$$
Let $\{(x_n,y_n)\}_n$ is a sequence in $A^C$ converging to $(x,y)$. We have $x_n y_n \le 0$, $\forall n$ $\implies \lim(x_n y_n) \le 0$, i.e. $xy \le 0$. Then, $(x,y) \in A^C$.
This shows that $A^C$ is closed, that's, $A$ is open.
A: If you can use continuous functions, then the set in question is the inverse image of the open interval $(0,\infty)$ under the continuous function $\mathbb R^2 \to \mathbb R$ with $(x,y)\mapsto xy$.
A: Draw the set in question. It has an easy-to-describe boundary, $B$; given any point $(x, y)$ in the set, it should visually be easy to find the distance from $(x, y)$ to $B$. Any radius smaller than this distance will work.
A: The region $xy>0$ is just the union of (1) the interior of the first quadrant, and (2) the interior of the third quadrant.  That is the set you're trying to prove to be open.
If $(x,y)$ is in the first quadrant, it distance from the $x$-axis is $y$ and its distance from the $y$-axis is $x$.  Make the radius of the disk less than or equal to both of those and then the disk won't intersect either of the boundaries, but rather will remain within the open set.  The same thing works in the third quadrant except that $x$ and $y$ are negative so you need $|x|$ and $|y|$.  So $|x|$ and $|y|$ do it, for both quadrants.
A: As you mentioned correctly one has to show, that for every $(x,y)\in M$ (I called the set M) one has to find a radius $r$ such that $B_r \subset M$.
So let $(x,y)\in M$ be arbitratry chosen, then if you choose $r=\min(|x|,|y|)/2$. $xy>0$ implies that $x$ and $y$ have the same signs. So wlog $x>0$ and $y>0$. Because of the choice of $r$ is $y-r>0$ and $x-r>0$.
That is why $M$ is open
A: It might also help to think about this visually. On the cartesian plane, your set represents quadrants $I$ and $III$, not including the $y$-axis or the $x$-axis. I recommend drawing this on paper, and then tossing a few points in quadrants $I$ and $III$. Next you'll want to figure out a way to guarantee that you can enclose the points in an open ball. One thing you can observe is that $(x,y)$ might be closer to the $y$-axis than it is to the $x$-axis. If it's closer to the $y$-axis, we know it's a distance of $|x|$ away from the $y$-axis, so an open ball of radius $|x|/2$ could be put around $(x,y)$ and still be contained in your set. On the other hand if the point is closer to the $x$-axis, an open ball of radius $|y|/2$ would suffice. In general you could use the ball $B_\varepsilon\left((x,y)\right)$ with $\varepsilon = \frac{1}{2}\min\{|x|,|y|\}$. Since your initial coordinate $(x,y)$ is arbitrary, you know you can put a ball around any point in the set  $\{(x,y)|xy>0\}$.
