Prove that $X=\left \{ (w,x,y,z) \in \mathbb{R}^{4} | w^2+z^2=1 \wedge x^2+y^2=1 \wedge wy-xz=0 \right \}$ is compact.

Question

Let $X=\left \{ (w,x,y,z) \in \mathbb{R}^{4} | w^2+z^2=1 \wedge x^2+y^2=1 \wedge wy-xz=0 \right \}$. Prove that $X$ is compact.

My attempt

As we are in $\mathbb{R}^{m}$, it's enough to prove that $X$ is bounded and closed. From the first two equations, we have tht $w,x,y,z \leq 1$. But I'm not sure how to prove that is closed, I mean, is it enough to say that all the conditions are equalitys?

Answer 1

Note that $f(x,y,z,w)=w^2+z^2$ ;$g(x,y,z,w)=x^2+y^2;h(x,y,z,w)=wy-zx$ are continuous functions and hence $f^{-1}\{1\},g^{-1}\{1\},h^{-1}\{0\}$ are closed sets.

$X=f^{-1}\{1\}\cap g^{-1}\{1\}\cap h^{-1}\{0\}$ is closed since intersection of closed sets is closed

Answer 2

To prove that it is indeed close, consider the functions $f,g,h : \mathbb{R}^4 \rightarrow \mathbb{R}$ given by $f(x,y,z,w) = x^2+y^2$, $g(x,y,z,w)=w^2+z^2$ and $h(x,y,z,w)=wy-xz$. Then $X = f^{-1}(1)\cap g^{-1}(1)\cap h^{-1}(0)$. Since $\{1\}$ and $\{0\}$ are closed in $\mathbb{R}$ and the three funcions are continuous, $X$ is the finite intersection of closed subsets hence it is closed.