Show that the set $S = \{(x, y) \in {\bf R}^2 \mid x > y\}$ is open So far I have done this: 
For any point $(x,y)$ in $S = \{(x, y) \in {\bf R}^2 \mid x > y\}$ you can take a neighbourhood radius $k$ which belongs to $S$, where $k = x-y$.
How do I prove that if $(a,b)$ belongs the the $k$-neighbourhood if $(x,y)$, $(a,b)$ belong to $S$.  
 A: Another way: the function $f\colon (x,y)\mapsto (x-y)\colon \Bbb R^2\to \Bbb R$ is continuous, and $(0,+\infty)$ is open, so its inverse image $f^{-1}[(0,+\infty)]$ is open in in $\Bbb R^2$. But that inverse image is just $\{(x,y)\in \Bbb R^2\mid x>y\}$.
A: Hint: Take a point $p \in S$. Let $d$ be the distance of $p$ to the line $x=y$, which is the boundary of $S$. Use $d$ to find a disk centered at $p$ and totally contained in $S$.
A: Another way to show this is to see it as a preimage of $(0,\infty)$ by the function $(x,y)\mapsto x-y$.
A: "For any point (x,y) in S you can take a neighbourhood radius k which belongs to S, where k = x-y."  No, you can't.  For example, the distance from the point (1/2, 0) to the line y= x is $\sqrt{2}/4$, smaller than 1/2- 0= 1/2.
The set is defined as {(x, y)| x> y}.  The boundary of that set is the line y= x which has slope 1.  The line perpendicular to that line, through point (a, b), is  y= -(x- a)+ b= -x+ (a+ b).  That line intersects y= x where x= y= (a+ b)/2.  The distance from (a, b) to ((a+b)/2, (a+b)/2) is $\frac{2|a-b|}{2}$.  That is the radius you want for your neighborhood.
A: It works in any ordered space $(X,<)$, so where the topology is the order topology. 
Suppose $(p,q) \in \{(x,y) \in X \times X: x > y \}$, so that $p > q$. Suppose there is some $r \in X$ such that $q < r < p$ (like in the reals), then $r^- = \{x \in X: x < q \}$ is open, and so is $r^+ = \{x \in X: x > r \}$. Also, $O = r^+ \times r^-$ is open in $X \times X$ in the product topology, $(p,q) \in O$, and if $(u,v) \in O$, we know that $u > r > v$, so $(u,v) \in \{(x,y) \in X \times X: x > y \}$, so $O$ shows that $(p,q)$ is an interior point. 
If no such $r$ between $p$ and $q$ exists (so $x > q$ implies $x \ge p$ etc.), we use $(p,q) \in q^+ \times p^- \subseteq \{(x,y) \in X \times X: x > y \}$. This is also easy to check. 
