How to prove that S = {$(x,y)| y > x^2$} is open? I don't understand how to do it at all. My professor tried so patiently to explain it to me but I just don't get it.
Here is what he did:
Choose any point in S, say (a,b). The point has a neighborhood around it inside S. Take some point inside the circle (neighborhood) centered at (a,b), say $(a + \delta, b + \nu)$ for $|\delta| < 1$ and $|\nu| < 1$. Now we want to try to do something with this point in order to narrow down our choice of $\epsilon$, the radius of the neighborhood centered at (a,b).
So
$(b+\nu) > (a+\delta)^2$
$b+ \nu > a^2 + 2a\delta + {\delta}^2$
$b - a^2 > 2a\delta + \delta^2 - \nu$
$b - a^2 > (2a + \delta)\delta - \nu$
Then he does something like
$|b - a^2| > |(2a + \delta)\delta - \nu|$
$|b - a^2| > |(2a + 1)\delta - \nu|$
And then he does triangle inequalities on $|(2a + 1)\delta - \nu|$, does some other stuff to get $|(2a + 1)\delta - \nu| < ((2|a|+1)\epsilon + \epsilon)$. He says stuff about maintaining control over our $\delta$ and $\nu$. Then divides $|b-a^2|$ by $2|a|+2$ to obtain $\epsilon$. 
Can somebody please explain to me why he does what he does? He really tried to help me and I really tried to understand it but I just don't understand the reason why we pick something in the circle, how to trap $\epsilon$ and pretty much everything. If someone can lay it out step by step, I would be so grateful! By the way, I do know what the definition of an open set is: A set S is open if there is some neighborhood $ \epsilon > 0$ for every point in S.
Thanks a bunch!
 A: $$S=\{(x,y):y>x^2\}=\{(x,y):y-x^2>0\}
=\{(x,y):f^{-1}((0,\infty))\}$$ where $f(x,y))=y-x^2$
Since f is continuous $f^{-1}(G)$ is open when $G$ is open. Here, $G=(0,\infty).$ This proves your aim.
A: Given $P \in \mathcal{S}$, you want to show that there exists $\epsilon$ such that $\mathcal{B}_\epsilon(p) \subset \mathcal{S}$.
Your professor is taking the following approach.  Suppose such an $\epsilon$ exists, and without loss of generality suppose $\epsilon \leq 1$.  Take a point in $\mathcal{B}_\epsilon(P)$, and determine the value of $\epsilon$ that will force the point to lie in $\mathcal{S}$.
Such an approach may work, but it is easy to get lost in the computation.  Here is a more "visual" approach.

Imagine a horizontal line through $P$.  This line intersects the parabola at two points $Q$ and $R$.  Suppose $Q$ is closer to $P$.  Take some point between $P$ and $Q$, call it $M$.  Now imagine a vertical line through $M$ intersecting the parabola at $N$.  Consider the axis-aligned rectangle with center $P$ and corner $N$.  The interior of this rectangle is a neighborhood of $P$ lying in $\mathcal{S}$.

Alternatively, consider the squared-distance function for the parabola, $f(x) = (a-x)^2 + (b-x^2)^2$.  It is not hard to prove with a little calculus that this function has a global minimum $\delta_\min>0$.  You can then let $\epsilon = \sqrt{\delta_\min}$.
A: The set $S$ in question is the "inside" of the parabola $y=x^2$. In order to prove that $S$ is open we have to show that,  given any  point $P:=(a,b)\in S$, we can draw a small disk $D_\epsilon$ centered at $P$ which is still completely contained in $S$. The  radius $\epsilon>0$ of this disk will of course depend on the chosen point $P$.
Your professor tried to find an admissible $\epsilon$ doing some computations and chasing inequalities.
Here is a geometric approach:  From $P$ draw a vertical downwards until you hit the parabola in the point $A=(a,a^2)$, and draw a horizontal  hitting the parabola in the points $B_\pm=\bigl(\pm\sqrt{b},b\bigr)$. Note that all three points $A$, $B_\pm$ are different from $P$. A disk with center $P$ whose lower half is contained in the interior of the triangle $\triangle(B_-,A,B_+)$ will then be completely contained in $S$.
