Proving a subset of $\mathbb{R}^2$ is an open subset. So the question is proving that: $U:=\{(x,y)\in\mathbb{R}^2 : xy\ne0\}$ is an open subset.
My method for doing so is define an open disk where its centre is contained within U, and then prove that the disk is also contained within U. Hence proving that it is an open subset? I am having trouble specifically with formulating these ideas into formal mathematic notation for the proof. 
Any help would be greatly appreciated! 
 A: The set you are considering corresponds to the union of all points not on the axes. Think pictorally. Pick a point somewhere. How big can you make a ball around it so it doesn't contain any points on the axes?
A: Your idea is on the right track. Start with a point $(x,y)$ in $U$ and show that if $(x',y')$ is sufficiently close to $(x,y)$, then $(x',y')\in U$, i.e. $x'y' \ne 0$.
If $xy\ne 0$, then $x \ne 0$ and $y\ne 0$. Hence if $0 < r < \min\{|x|,|y|\}$, and $(x',y')\in B_r\big(\,(x,y)\,\big)$, show that $x'y' \ne 0$.
A: Another way would be like consider the continuous mapping $\varphi:(x,y)\rightarrow xy$ on ${\bf{R}}^{2}$, and the set in question is $\varphi^{-1}(-\infty,0)\cup\varphi^{-1}(0,\infty)$ which is open.
A: Note that the complement $C= \{ (x,y) | xy = 0 \} = (\{0\} \times \mathbb{R} ) \cup ( \mathbb{R} \times \{0\})$ is closed.
A: Here's yet another proof.
Each half plane is open somewhat trivially - any ball with radius less than whatever coordinate is relevant will do. Now the union of any pair of half planes is open, since the union of opens is open. So each pair (left/right and top/bottom) is open. The intersection of these two pairs is your set, and the intersection of two open sets is open.
Of course, my initial remark extends readily to give a direct proof, but hey, whatever gets the job done.
