Q:

Show that $D=\{(x,y) \in R^2:-1<x<1,0<y<\sqrt{1-x^2}\}$ is open

I got this far. $D$ is the half open circle in the plane with radius 1. Let $x \in D$ and take $\epsilon=\text{min}\{1-||x||,|y|\}$ where $y$ is the second component of $x$. Let $z \in B(x,\epsilon)$ how to show that $z$ is indeed in $D$?

A:

Define $r=\min\{ \sqrt{1-\Vert x \Vert^2}, x_2  \}$,  you need to show that $B(x,r)\subset D$. i.e, for all $(z_1,z_2)\in B(x,r)$ we have that:
$z_2\geq 0$ and $z_1^2+z_2^2<1$
Notice then that $\vert x_1-z_1\vert<r$ and $\vert x_2 -z_2\vert<r$. 
You can conclude that:
$z_2=x_2+(z_2-x_2)\geq x_2 -r \geq x_2-x_2=0$ since $x_2>0$ as $x\in D$.
Also:
$z_1^2+z_2^2= x_1^2 +x_2^2 +\big( z_1^2-x_1^2 \big)+ \big( z_2^2-x_2^2 \big)=\Vert x\Vert^2+\Vert z-x\Vert^2< \Vert x\Vert^2+r^2\leq $
$\leq \Vert x\Vert^2 + 1-\Vert x\Vert^2=1$

