The problem is to prove that the set $A=\{(x,y)\in\mathbb{R}^2: x^2 + y^2 > 4, y < 6\}$ is open in $\mathbb{R}^2$
My attempt:
Let $S=(0,0)$
Let's take an arbitrary $x=(x_1,x_2) \in A$ and define $R = \min\{\lvert{d(x,S)-2}\rvert,\lvert x_2-6 \rvert \}$, so we have an open ball $K(x,R)$ To prove that the set is open we need to prove that for an arbitrary $y=(y_1,y_2) \in K(x,R)$ that $y$ has to be in $A$.
By a few manipulations, mainly using the triangle inequality and the points $S,x,y$ I've managed to prove that $y_1^2+y_2^2 > 4$.
But proving $y_2 < 6$ has proven to be very difficult and I'm left without ideas as to what to do specifically.I have a feeling I'm missing something much simpler here.
Thanks in advance!