I am working on a problem in which I must determine if the extrema (I found these in the initial part of the problem) of a function $f: \mathbb{R}^3 \to \mathbb{R}$, constrained to the set
$$C = \{(x, y, z): x^2 + y^2 + z^2 = \pi\},$$
are global extrema.
NB: $C$ is the sphere of radius $\sqrt\pi$, centred at the origin.
I know that if $C$ is compact (both closed and bounded), then the extrema of $f$ constrained to $C$ (found in the initial part of the problem) will be global extrema.
I think that $C$ is bounded, but am not sure if my understanding of a bounded set in $\mathbb{R}^3$ is correct, especially after having read this.
Also, I am not sure how to determine if $C$ is closed.
The definition of a closed set is one whose complement is open (or a set which contains all its limit points).
I am not sure if the complement of $C$ is open, seeing as some points in the complement of $C$ lie within the sphere and others lie outside the sphere.
I would appreciate some help!
$g(x,y,z)=x^2+y2+z^2$ is continuous therefore $g^{-1}(\pi)=C$ is closed
$C$ is not empty $(\sqrt{\pi},0,0)$
$\forall (x,y,z)\in C, |x| = \sqrt{\pi-y^2-z^2} \le \sqrt{\pi},|y| = \sqrt{\pi-x^2-z^2} \le \sqrt{\pi},|z| = \sqrt{\pi-y^2-x^2} \le \sqrt{\pi}$
So $C$ is bounded
Therefore C is compact