Are the following sets compact:
- $A=\{(x,y,z):x^2+2y^2-3z^2=1\}$
- $B=\{(x,y,z):|x|+2|y|-3|z|\le1\}$
$A=f^{-1}\{1\}$ where $f(x,y,z)=x^2+2y^2-3z^2$ is a continuous function and hence $A$ is closed.
However $A$ is not bounded as $(n,n,\sqrt{\dfrac{3n^2-1}{3}})\in A$ which is unbounded for large $n$.
$B=g^{-1}\{(-\infty,1]\}$ where $g(x,y,z)=|x|+2|y|-3|z|$ is a continuous function and hence $B$ is closed.
Now $B_1=\{(x,y,z):|x|+2|y|-3|z|=1\}\subseteq B$.
However $B_1$ is not bounded as $(n,n,{\dfrac{3n-1}{3}})\in B_1$ which is unbounded for large $n$.
Hence $B$ is not bounded.
But the answer is given $B$ is compact..Please help.