Prove that $S= \{ (x,y) : x^2 - y^2 <1 \}$ is open in $\mathbb{R}^2$ 
Prove that $S= \{ (x,y) : x^2 - y^2 < 1 \}$ is open in
  $\mathbb{R}^2.$

The question itself is rather easy and trivial by observing the $S$ in $\mathbb{R}^2$ geometrically. But if we are restricted to prove the above by using the definition of open set ONLY, how should we write down the proof formally?
A set $P$ is called open if for every $x \in P$, there exists $r>0$ such that $B(x, r) \subset P$.
Particularly, in this question, what value of $r$ should we choose for a given $(x,y) \in S$?
Thanks in advance.
 A: Consider the function $f(x,y) = x^2 - y^2$, $f\colon \mathbb{R}^2 \to \mathbb{R}$. It is certainly continuous. Then $S = f^{-1}( (-\infty, 1) )$. Since $(-\infty, 1)$ is open, $S$ is open too.
As for your question about proving openness from the very definition, note that the boundary of $S$ is just a hyperbola. For any point in $S$ you have enough room to find a ball around that point between the branches of this hyperbola.
A: Calculating the exact distance between a disc and the given set can indeed be iffy since a fourth order equation appears, but fortunately it is also unnecessary. Instead, observe that a disc of radius $a$ is a subset of a square with sides $2a$. Also, the upper branch of the border $\partial S$ is an increasing function. Therefore, given $(x,y>0)\in S$, take a disc with radius $r=\min\left(y,(y-x+1)/2\right)$ (this follows from the consideration of the square). You will have to check that $r>0$. For this, note that the assumption that $x\geq y-1$ leads to $(x,y)\not \in S$.
For $y=0$, use $r=(1-x)/2$ instead.
A: In general, using directly the definition of open set is rather complicated. The easiest way to prove that $S$ is open is noting that it is the preimage of an open set, given a continuous function. (As it is detailed in other answer).
In this example, you can rely on geometric background to state that for every point $X(x,y)$ in $S$, there is some tangent circle of radius $r_X$ to the hyperbola $x^2-y^2=1$. Take for every point $X$ a ball of radius $r_X/2$.
