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

$$\mathbf {The \ Problem \ is}:$$ 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 $$\mathbb R^n .$$

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.
• Sir, for $n = 1$, the set defined by $x^2+y^2-1 =0$ is an uncountable algebraic set .And, how to answer the question for uncountable $S$ and where to use the fact "algebraic" ??? Oct 22 '19 at 14:27
• @RabiKumarChakraborty I used the fact $S$ is a countable union of algebraic sets in two places: one of them is algebraic sets are algebraic again under the linear transformation and restriction to a plane. The other is every algebraic subset of $\mathbb{C}$ is finite. Oct 22 '19 at 14:54
• O, thanks Sir, for rectifying my mistake, and Sir can you provide some hints about the non trivial case ,when $S$ is uncountable ??? Oct 22 '19 at 16:57
• In that case, $S$ could be $\mathbb{C}^n$ itself. (For example, $\mathbb{C}$ is a union of singletons.) I have no idea when $S$ is a union of $\kappa$ algebraic sets, where $\aleph_0<\kappa<2^{\aleph_0}$, and it seems to be related to deeper set theory. Oct 22 '19 at 17:12