Review for Exam 1 Problem 6

In Problem 6 of the Exam 1 Review sheet, we are asked to state whether the sets $R={z\in\mathbb{C}:1<|z|<2,0\leq \arg(z)<\pi}$ and $R^2$ are closed, open, or neither. We can show that a set is not closed by showing there exists a boundary point of the set not contained in the set. By this reasoning, $R$ and $R^2$ cannot be closed since $-1$ is a boundary point of both sets that is contained in neither set.

Please assume I know what $R$ and $R^2$ look like in the complex plane. I need to know how to show a set is neither open nor closed, and whether $R$ and $R^2$ fit this criteria.

Perhaps the first thing we should try to do is sketch an image. I guess that $R$ looks something like so:

The colored edges and white dots don't signify anything specific, other than the fact that those are the components of the boundary that we need to consider. For reach such edge, you should ask if it's in or out of the the set. To show that a set is not closed, you should find a point (probably on the boundary) with the property that every disk you draw about that point contains points both inside and outside of the set. To show that a set is closed, you should show that there is no such point. That is, for every point in the set, you need to show that there is a disk about that point that is wholly contained in the set.

I guess you need to come up with a picture of $R^2$ and do the same things there.