Determine function's domain type

Question

I'm trying to solve this two excercise:
1) Determine the domain of $f(x,y)=\sqrt{3-xy}$, then say if it is open, close, not open neither close; say if it is bounded or not.

To find the solution I solve $3-xy \ge 0$ and I find $y \le \frac{3}{x}$. In my opinion the domain is not open neither close, and it is not bounded.


2) Determine the domain of $f(x,y)=\log({3-xy})$, then say if it is open, close, not open neither close; say if it is bounded or not.

To find the solution I solve $3-xy \gt 0$ and I find $y \lt \frac{3}{x}$. In my opinion the domain is open, and it is not bounded.

I'm not sure of my answers can someone help me?

Answer 1

1) Be careful when dividing by $x$. If $x$ is negative, the inequality switches directions. If $x$ is $0$, you can't divide by it, but $x=0$ is still a solution to your inequality. The domain is slightly more complicated than what you have expressed.

2) Ditto for the dividing by $x$ problem. Also, depending on what you come up with for an answer, it might not be open EVEN IF you have strict inequalities $<$ running around everywhere. That gives you a good hunch that it might be open, but look carefully at the boundary. Sets don't necessarily have to be open or closed.

In either case, it's important to justify why the set is open or closed. Go back to whichever definition your book uses and see if these domains have those properties.

Answer 2

You want to solve $xy \le 3$.

$x = 0$ is a solution for all $y$.

If $x > 0$, then $y \le \dfrac 3x$.

If $x < 0$, then $y \ge \dfrac 3x$

Below is a graph of $xy \le 3$. clearly its complement is an open set. So it must be a closed set.