Show a set is open in C

Question

Show that the set O={z$\in\Bbb C$:Re(z)<0, Im(z)>0} is open in $\mathbb C$

I know this is the second quadrant, but I don't know how to do the proof.

Answer 1

Suppose that $z \in O$. We need to find $\epsilon > 0$ such that $w \in O$ for all $w$ with $\left|w-z\right| < \epsilon$

Let $z = -a + bi$, where $a,b$ are positive reals.

Then just take $\epsilon = \text{min}(a,b)$. For then, if $\left|w-z\right| < \epsilon$, where $w = -c + di$,we have

$$\left|a-c\right| + \left|b-d\right| \leq \left|w-z\right| < \epsilon$$

and so it follows that $c, d > 0$, that is $w \in O$.

Answer 2

Let $z_{0}=x_{0}+iy_{0}$ ($x_{0}$ and $y_{0}$ real numbers) be in this quadrant. Consider the ball $D(z_{0},r_{0})$ where $r_{0}=\min(|y_{0}|,|x_{0}|)$. Assume that $ z=x+iy (x,y \in \mathbb{R}) \in D(z_{0},r_{0})$. Then $|z-z_{0}|=\sqrt{(x-x_{0})^{2}+(y-y_{0})^{2}}<r_{0}$. Hence $|x-x_{0}|<|x_{0}|$ hence $x_{0}<x-x_{0}<-x_{0}$, [$x_{0}<0$ so $|x_{0}|=-x_{0}$], hence $2x_{0}<x<0$ so $x<0$. Similarly, $|y-y_{0}| \leq |z-z_{0}|<|y_{0}|$ so $-y_{0}<y-y_{0}<y_{0}$, hence $0<y<2y_{0}$.