A problem on the connectedness relating to an algebraic subset of $\mathbb C^n$

Question

$\mathbf {The \ Problem \ is}:$ If $S$ is a countable union of algebraic subsets of $\mathbb C^n$, i.e. a set defined by algebraic equations in the coordinates of $\mathbb C^n$ then show that $P= \mathbb C^n \setminus S$ is poly-line connected i.e. whose any two points can be joined by finitely many broken lines lying in that set .

$\mathbf {My \ approach} :$ Actually,I tried in this way that for any two point $p,q \in P$, we join them by a line segment $L$ and as $S$ is closed, we let two open balls $N_p$ and $N_q$ of $p$ and $q$ such that both of them doesn't meet $S .$

Then for any point $r$ on $L$, not in $S$, we let such a ball around $r$, not meeting $S$ and for points in $S$, we let a $\delta$ ball around them, they cover the connected set $L$, then I think I need to use the finite chain condition on connected sets and poly-line connectedness of $\mathbb C^n \setminus K$ where $K$ is countable .

But, I can't think of the fact where to use the "algebraic set" .

I found a result that such sets are nowhere dense in $\mathbb R^n .$

Answer 1

We can use induction on $n$.

We first assume that $\mathbb{C}^n$ minus a countable union of algebraic sets is countable. Let $a,b \in \mathbb{C}^{n+1}\setminus S$, where $S$ is a countable union of algebraic sets $S_k$ of $\mathbb{C}^{n+1}$. Consider any hyperplane $H$ containing $a$ and $b$. Then $(\mathbb{C}^{n+1}\setminus S)\cap H = H\setminus(S\cap H)$ is homeomorphic to $\mathbb{C}^n$ minus some countable union of algebraic sets via linear transformation $T$. By the inductive hypothesis, we can find a poly-line path between $T(a)$ and $T(b)$. Therefore, we can find a poly-line path between $a$ and $b$.

For the case $n=1$, every algebraic subset of $\mathbb{C}$ is just a finite set of points. Hence, a countable union of algebraic sets is countable (including finite sets, of course.) It remains to show that $\mathbb{C}\setminus S$ is poly-line connected for any countable $S$, but there are other answers proving this statement: for example, you can see an answer by Brian M. Scott.