I have subset
$$D = \{z: \text{Re}(z) > 2, \text{Im}(z) \leq1\}$$
So i think that set is not open because contain boundary points, but how prove this?
Let's say it's open. so by defenition we have: For all points of the set, they are contained together with some ball. Let's take an arbitrary point (with the property $\text{Im}(z_0)=1$),
$$z_0 \in D$$
Let's look at the ball with the center at our point.
$$B_r(z_0) = \{z: |z – z_0|<r\}$$
Where $r>0$ is the positive number. So…
I have a problem with building a contradiction. Intuition tells us that if a number
$r$
has $\text{Im}(r) > 0$
So we have $z_o + r = c \notin D$
is it close?
Help me develop intuition and technique for this kind of task. (prooving subset of complex plane is open or close)
Best Answer
Your are in the right way to prove $D$ is not open. It suffices to take $ \ z=z_{0}+\frac{i}{2}r \ $, then $ \ |z-z_{0}|=\frac{r}{2}<r \ $ and $ \Im(z)=1+\frac{r}{2}>1. $
$A \subset \mathbb{C} \ $ is closed if and only if the limit of every convergent sequence $ (z_{n})_{n} \subset A $ is also in $A$.
Take $ \ z_{n}=2+\frac{1}{n} \ $, then $z_{n}\in D$ and $ \lim_{n\to \infty}z_{n}=2 \notin D $. Hence $D$ is not closed.
Note that the only subsets of $\mathbb{C}$ which are both open and closed are $ \emptyset$ and $ \mathbb{C} $, since $\mathbb{C}$ is connected.