Prove that the following sets are open:
i) $A=\{ (x_1,x_2) \ | \ x_1^2 + x_2^2 < 1 \ \text{and} \ x_2> 0\}$
ii) $B=\{ (x_1,x_2) \ |\ x_1\neq 0\}$
My attempt:
i)
I was thinking that I want to get half the distance between $(x_1,x_2)$ and $(x_1, \sqrt{1-x_1^2})$ and half the distance between $(x_1,x_2)$ and $(\sqrt{1-x_2^2},x_2)$, and then take the minimum of those two numbers to be the radius of my ball.
After a very long time, I noticed that when $(x_1,x_2)$ is in the yellow area, it will be in the ball $B_r (x_1,x_2)$, where $r=\frac{-x_1 + \sqrt{1-x_2 ^2}}{2} $. And when $(x_1,x_2)$ is in the green area, it will be in the ball $B_r (x_1,x_2)$, where $r= \frac{-x_2 + \sqrt{1-x_1^2}}{2} $. So, when we take $r=\min \{\frac{-x_1 + \sqrt{1-x_2 ^2}}{2}, \frac{-x_2 + \sqrt{1-x_1^2}}{2} \} $, we will have an open ball $B_r (x_1,x_2)$ that is contained in $A$.
What do you think?
ii) Let $(x_1,x_2) \in B$, then we have $(x_1,x_2) \in B_r(x_1,x_2)$, where $r=\frac{|x_1|}{2}$. And I draw it as the following:
Are my answers correct? Thank you!