Is the set $A = \{(x,y,z) \in \mathbb{R}^3 : 0 < x^2 +y^2 \le 9 \}$ open or closed? Consider $A = \{(x,y,z) \in \mathbb{R}^3  :  0 < x^2 +y^2 \le 9 \}$.  Is this set open, closed, and why? The origin is excluded so I guess that it is an open set, what about its boundary?  Is this set bounded?
 A: $A$ is neither open nor closed, to see this note that

*

*$\big((1/(n+1), 0, 0)\big)_{n \in \mathbb{N}}$ is a sequence in $A$ that has limit $(0,0) \notin A$ so $A$ is not closed

*no nieghborhood of the point $(3,0,0)$ is entirely contained in $A$, so $A$ is not open.

You may want to know that despite the terminology there are subsets that are neither open nor closed and subsets that are both open and closed (most notably the entire space and $\emptyset$ are always both open and closed).
A: $$Int(A) = \{(x,y,z) \in \mathbb{R}^3  :  0 < x^2 +y^2 \color{red}< 9\} \ne A.$$
$$Cl(A) \  = \{(x,y,z) \in \mathbb{R}^3  :  0 \color{red}\le x^2 +y^2 \le 9\} \ne A.$$
A: A set is closed if and only if it contains its limit points. Notice that $0$ is a limit point of $A$, hence $A$ cannot be closed since $0 \notin A$.
A set is open if and only if it is disjoint from its boundary. Note $A$ contains boundary points (the circle of radius $3$) so cannot be open.
To determine the boundary one might draw a picture, you can then easily verify that the boundary is $\{(x,y,z) \: \vert \: x^2 + y^2 = 9\} \cup \{(0,0,z)\}$.
