TITLE: Prove that subset of complex plane is open or is closed (or both or neither )
QUESTION [0 upvotes]: 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)

REPLY [0 votes]: We can use the definition of open sets in metric spaces. $D$ isn't open because a disk centered on $i$ will always have a point outside $D$ so it cannot be entirely contained in $D$. $^cD$ (the complement of D) isn't open either because a disk centered on $2$ will always have a point outside $D$. Hence $D$ is neither open nor closed.  
This is not a detailed proof but it's really easy to show an example point (depending on the radius $r$ of the disk) in both cases : $(1+r/2)i$, $2+r/2$.
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