Classify domain in $\Bbb R^3$ as open/closed/bounded/connected If I have a function:
$$f(x,y,z) = \frac{{\sqrt {\ln (1 + {z^2})} }}{{\left| x \right| + \left| y \right| + \sqrt {xy - 1} }}$$
And I need to classify its domain as open/closed/bounded/connected.
My thoughts:
The only restrictions are that xy>1, so the function is bounded. Since it doesnt contain its boundary points, it is not closed. And I think it is also open? How do I tell if the domain is connected?

You are right. $xy \gt 1$ is the only required stipulation. $f$ is defined everywhere else. 

This is the graph for $xy = 1$ in $\Bbb R^3$. Any values below(or above as you may picture it) the graph (or the two hoods ) satisfies $xy \gt 1$. So $D(f)$ is not bounded. Since given any $M \gt 0$ there is $(x, y, z) \in D(f)$ such that $ x, y, z \gt \text{Max}\{M, 1\}  \implies |(x, y, z)| \gt M $. It can be similarly proven that $D(f)$ is not bounded below. You are right about it being open. But maybe a little more justification is needed. It is clearly not connected. I do not know your definition for connectedness. According to mine $D(f)$ is disconnected since there are two open sets $A = \{(x, y, z) \ | \ x \gt 0 \land y \gt 0 \}$ and $B = \{(x, y, z) \ | \ x \lt 0 \land y \lt 0 \}$ such that $A \cap D(f)$ and $B \cap D(f)$ are disjoint, non-empty and have union $D(f)$. 
