How to Show that $S=\{(x,y)\in R^2:x>y^2\}$ is open Show that $S=\{(x,y)\in R^2:x&gt;y^2\}$ is open

quite simple one. We need to choose $\epsilon$ for open balls: $D((x_0,y_o),\epsilon)\subset S$ ,$\forall x_o,y_o\in S$.

we can take $\epsilon$ as the closest distance to the parabola. but couldnt do
You don't need to take $\epsilon$ as the closest distance to the parabola.  

Take $(x,y) \in S$.  We have $x &gt; y^2 \implies x - y^2 &gt; 0$.

If $0 &lt; r &lt; 1$, note that
$$
(x-r) - (y+r)^2 = (x - y^2) - (r - 2yr + r^2) &gt; (x - y^2) - (2 + 2|y|)r
$$
So, select and $r&gt;0$ satisfying $r &lt; \min\{(x - y^2)/(2 + 2|y|),1\}$.  How can we show that the ball of radius $r$ about $(x,y)$ is contained in $S$?
Property: A set is open if its complement is closed. Every open ball of radius d around any point of x&lt;=y² intersects infinite points of x&lt;=y² so the complement of S is closed and S is open.