Compact surface

Question

To check if the surface $x^2-y^2+z^4=1$ is compact, we have to check if the surface is closed and bounded.

Could you give me some hints how exactly we check that?

How can we check if it closed and how if it is bounded?

Answer 1

It's not bounded. Fix $z=1$, then you have a subset of the surface defined by $x^2-y^2 = 0$, so both $x$ and $y$ can go off to $\infty$ without any constraint (besides being equal up to sign).

So the surface is not compact. To check nonetheless whether it's closed look at a sequence of points satisfying the equation and assume the sequence converges. If the limit also satisfies the equation then the surface is closed (this is true, since you are dealing with continuous functions).

Answer 2

As the first answer indicates, this set is not bounded.

To determine whether the surface is closed, instead of thinking about convergent sequences of points on the surface, think about the function defining the surface. Let $f(x,y,z)=x^2-y^2+z^4$ and notice that the surface is the set $f^{-1}(\{1\})$, i.e. the set $\{(x,y,z)~:~x^2-y^2+z^4=1\}$. Since $f$ is a continuous function and $\{1\}$ is a closed set, $f^{-1}(\{1\})$ is closed.

Answer 3

Unfortunately, this will not be bounded. Note that $z=1$ will always satisfy $z^4 = 1$ so $x^2 - y^2 = 0$ will suffice, i.e. just take $x = y$ and now you have infinitely many solutions as arbitrarily large as you wish.

The points on this surface are $$\{(x,y,z) \in \mathbf{R}^3 \mid f(x,y,z) = x^2 - y^2 + z^4 - 1 = 0\} = f^{-1} (0)$$.

Now it is clear that it is closed since it is the inverse image of a closed set under a continuous map. I want to emphasize this technique which can be generalized much further, such as to show any zero set of a continuous function is closed. These types of problems frequently appear on examinations and qualifiers (e.g. Berkeley).