Sphere in $C^3$ is not compact? Let $A=\{(x,y,z) \in \mathbb{C}^3 \mid x^2+y^2+z^2=1\}$ is not compact with Euclidean topology?
My attempt — it is not bounded since we can choose $x$ and $y$ to be arbitrarily large and this forces value of $z$ which is in $\mathbb C$ because it is algebraically closed.
 A: That's right! Maybe a slightly cleaner way to say the same argument it would be, "Since $\mathbb{C}$ is algebraically closed, for any $a, b\in\mathbb{C}$ the polynomial $P(z)=a^2+b^2+z^2-1$ has a root; so for every $a, b$ there is some $c$ such that $(a,b,c)\in A$, which means $A$ is not bounded and hence not compact."
(Depending on what level of detail is wanted, you may also need to justify the conclusion that $A$ is not bounded. But this isn't hard: just show that $A$ contains points arbitrarily far from the origin, and no bounded set contains points arbitrarily far from the origin.)
A slightly different spin you can put on the same idea is: you've shown that since $\mathbb{C}$ is algebraically closed, the projection of $A$ onto $\mathbb{C}^2$ - the set $\{(x,y)\in\mathbb{C}^2: \exists z((x,y,z)\in A)\}$ - is all of $\mathbb{C}^2$, and hence not compact. You can now conclude that $A$ itself is not compact from the following two facts (which are themselves important):


*

*The projection map $(x,y,z)\mapsto(x,y)$ is continuous.

*The continuous image of a compact set is always compact (so contrapositively, if a set has a non-compact set as a continuous image, then the original set is also non-compact).
This is overly abstract, certainly, but it's a good "toy example" of how to apply general principles about the behavior of topological properties (in this case, how compactness and continuity interact).
A: Yes, it is not bounded. But there is no need to mention $z$ or the fact that $\mathbb C$ is algebraically closed. Just note that $A$ contains all vectors of the form $\left(ix,\sqrt{1+x^2},0\right)$, with $x\in\mathbb{R}$. And$$\left\|\left(ix,\sqrt{1+x^2},0\right)\right\|=\sqrt{1+2x^2}.$$
