What exactly is the difference between a closed set and a set that is neither closed nor open?

Question

So, my textbook says that a set is said to be closed if it contains all the boundary points. As for the definition of a set that is neither closed nor open, it says that the set looks somewhat like a punctured disk. They gave this set as an example: 0 < |z| $\le$ 1. What I didn't understand is, this set still contains all the boundary points. So, why was this considered to be a set that is neither open nor closed?

Answer 1

In your example $0$ is a boundary point: the sequence $1/n$ converges to $0$, while the latter point isn't in the punctured disk. So the punctured disk isn't closed.

Answer 2

Reading the definition like a normal English sentence, it may seem like a closed set needs to have just boundary points. But with math definitions, what they're often saying is an extra property or constraint that the object needs to have.

The definition,

a set is said to be closed if it contains all the boundary points

can be alternatively read as

If you have a set, and that set contains all of its boundary points (in addition to the normal points of the set), then we call it a closed set

A different definition that I learnt and prefer is

a set is said to be closed it it contains all of its limit points

A limit point is the limit of a sequence that uses only points from the set. In many cases, the limit is already inside the set (i.e. an interior point), but not always. For example, if you start from inside the set, and move towards the outside, the limit of this "sequence" will be a boundary point. Another reason why I prefer the limit point definition is your example. $0$ may not intuitively feel like a boundary point for the punctured point, but maybe it's easier to see it as a limit point; that is, you can create a sequence in the set whose limit is the $0$ point. But because $0$ is not included in the set, the set is not closed, even though it contains all the limits points on the outer boundary ($|z|=1$).