Let $K = \{(x,y,z)\in\mathbb{R}^3 : x^2 + y^2 = z^2(1-z^2)\}.$ Determine with proof whether $K$ is complete, compact, and/or connected using the Euclidean metric.
I think $K$ is complete and compact, and to show this I use the fact that $A$ is the inverse image of the continuous function $f(x,y,z) = x^2 + y^2 – z^2(1-z^2)$ and that $K$ is bounded (by finding the maximum of the function $g(x,y,z) = ||(x,y,z)||^2$ which is $1$). I'm not sure how to show that $K$ is connected. I think it is, and I tried defining a (continuous) path on $K$, but I wasn't able to find such a function. I know that to show it's not connected, I need to find a pair of open sets that separate $K$, but I wasn't able to find such sets either.
Best Answer
Yes, $K$ is bounded. Note that if $(x,y,z)\in K$, then $z\in[-1,1]$; otherwise, $x^2+y^2=z^2(1-z^2)<0$. But then\begin{align}\|(x,y,z)\|&=\sqrt{x^2+y^2+z^2}\\&\leqslant\sqrt{z^2(1-z^2)+z^2}\\&=|z|\sqrt{2-z^2}\\&\leqslant\sqrt2.\end{align}And $K$ is closed, since $K=f^{-1}\bigl(\{0\}\bigr)$, with $f(x,y,z)=x^2+y^2-z^2(1-z^2)$. Thus, $K$ is compact.
In order to prove that $K$ is connected, consider the set$$A=\left\{(x,y)\in\Bbb R\times[-1,1]\,\middle|\,|x|\leqslant|y|\sqrt{1-y^2}\right\}.$$It's a connected set. More generaly, for any function $f\colon[a,b]\longrightarrow[0,\infty)$, the set$$A_f=\left\{(x,y)\in\Bbb R\times[a,b]\,\middle|\,|x|\leqslant f(y)\right\}$$is connected. Actually, it's path-connected: if $(c_1,d_1),(c_2,d_2)\in A_f$ then you can construct a path from $(c_1,d_1)$ to $(c_2,d_2)$ in $A_f$ as follows:
Now, consider the set$$K_0=\left\{\left(0,|z|\sqrt{1-z^2},z\right)\,\middle|\,z\in[-1,1]\right\}.$$It is a subset of $K$ and it is not empty, since, for instance, $(0,0,0)\in K_0$. Now, let$$K_+=\left\{\left(\sqrt{z^2(1-z^2)-y^2},y,z\right)\,\middle|\,(y,z)\in A\right\}$$and let$$K_-=\left\{\left(-\sqrt{z^2(1-z^2)-y^2},y,z\right)\,\middle|\,(y,z)\in A\right\}.$$Both are connected, since both of them are the image of $A$ with respect to a continuous map. But $K=K_+\cup K_-$ and $K_+\cap K_-=K_0\ne\emptyset$. But then $K$ is connected, since it's the union of two connected sets with non-empty intersection.