Proving a set is open 
Show that the set of $\mathbb{R}^2$ given by $E = \{(x, y) \in
 \mathbb{R}^2 : x > y\}$ is open.

I know I must determine a radius $r$ such that $$\{(x_1, y_2)\in \mathbb{R}^2: \sqrt{|x-y|^2 +|x_1-y_1|^2}<r\}$$
but how can I find such an $r$?
 A: One alternate definition I like about open and closed sets is based on their borders. Define the border of a set $S$ as "the set of all points $B$ s.t. if $x \in B$, for every neighborhood $U$ around x, at least point in U is a point in $B$ and one point is in the compliment $B^c$". 
From this it's pretty easy to get working definitions of open and closed sets: Clearly every point in $B$ is a limit point of $S$, and every limit point of $S$ will either be in S or be in $B$ (if it's not, then there are neighborhoods without any points in $S$, so it can't be a limit point). So a closed set is one which contains its border. Similarly, given a point x, if we can find an open ball $R$ around x st $R \subset S$, x can't be on the border. So an open set is one which does not contain any points in its border.
Pulling back to your problem. What's the border of $E$? It shouldn't be hard to argue that it's $B_E = \{(x,y): x=y \}$. Now, is $B_e \cap E = \emptyset$? If $x>y$, then $x\neq y$ and $B_e \cap E = \emptyset$. Therefore E is open.
EDIT: You'll probably want to do it a more conventional way too, but hopefully this is enough to convince you that it's true.
A: Hint: Locate the region of the plane where $x\gt y$.  Put a point $P=(x_1,y_1)$ in it, say with $P$ fairly close to the line $y=x$.  
Now move to the left from $P$ until you hit the line $y=x$, say at $Q$. Then $Q$ has coordinates $(y_1,y_1)$.
Let $x_1=y_1+\delta$, where $\delta$ is positive. Argue that the distance from $P$ to the line is $\frac{\delta}{\sqrt{2}}$. So if we take a disk with centre $P$ and $\dots$. 
A: If you know about continuous functions, then $E=f^{-1}(0,\infty)$ for $f:\mathbb R^2 \to \mathbb R$ given by $f(x,y)=x-y$. It follows that $E$ is open because $f$ is continuous and $(0,\infty)$ is open in $\mathbb R$.
