# Question regarding the path-connectedness of a set in $\mathbb C^n$

$$\mathbf {The \ Problem \ is}:$$ If $$X$$ is a countable union of algebraic subsets of $$\mathbb C^n$$, i.e. a set defined by algebraic equations in the coordinates of $$\mathbb C^n$$(under standard topology) then show that $$P= \mathbb C^n \setminus X$$ 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 $$L$$, the line segment $$L_{pq}$$ is connected and as $$X$$ is closed, we let two open balls $$N_p$$ and $$N_q$$ of $$p$$ and $$q$$ such that both of them doesn't meet $$X .$$

Then for any point on $$L_{pq}$$, not in $$X$$, we let such a neighborhood not meeting $$X$$ and for points in $$X$$, we let an $$\epsilon$$ neighborhood, they cover the connected set $$L_{pq}$$, then I think I need to use the finite chain criterion of 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 $$\mathbb R^n .$$

Let $$X=\bigcup_k X_k$$, where each $$X_k\subset\mathbb C^n$$ is a complex algebraic subset. For any $$p,q\in P$$ the complex line $$L_{pq}$$ is not contained in $$X_k$$, hence $$X_k\cap L_{pq}$$ is a proper algebraic subset of $$L_{pq}$$, hence a finite set of points. Thus $$X\cap L_{pq}$$ is countable. Now, the complex line $$L_{pq}\subset\mathbb C^n$$ is a real plane $$H\subset\mathbb R^m$$, $$m=2n$$, and we can repeat an argument given somewhere in mathstackexchange: pick a real line $$\ell\subset H$$ through $$p$$ that does not meet $$H\cap X$$ (possible because this set is countable); next pick another $$\ell'\subset H$$ through $$q$$ that does not meet $$H\cap X$$ and is not parallel to $$\ell$$. Then the intersection point $$x\in\ell\cap\ell'$$ gives a polygonal path $$\gamma$$ from $$p$$ to $$x$$ in $$\ell$$, then from $$x$$ to $$q$$ in $$\ell'$$, which does not meet $$X$$. We are done.
This is an instance of a more general fact: if $$X\subset\mathbb R^m$$ is a countable union of smooth submanifolds $$M_i\subset\mathbb R^m$$ of codimension $$\ge2$$, then $$P=\mathbb R^m\setminus X$$ is connected by polygonal paths.
Indeed, every complex algebraic set $$X_k$$ in $$X=\bigcup_k X_k$$ is a union of complex analytic submanifolds of $${\mathbb C}^n\equiv{\mathbb R}^m$$, which have an underlying structure of real smooth submanifolds. Namely, regular locus, then regular locus of the singular locus, then regular locus of the singular-singular locus... The topological real codimension of those manifolds is at least $$2$$. Thus we have that $$X\subset{\mathbb R}^m$$ is a countable union of smooth manifolds of codimension at least $$2$$. In passing, note this works the same for complex analytic subsets $$X_k$$ of $${\mathbb C}^n$$, each one being a countable union of complex analytic submanifolds.
Now let us see that $$P=\mathbb R^m\setminus X$$ is connected by polygonal paths. Pick any two points $$p,q\in Y$$. Then every plane $$H$$ containing both points is given by a third point $$c\in\mathbb R^m$$, and by an standard transversality argument, the $$c$$ can be chosen for $$H$$ to be in general position with all $$M_i$$. Indeed, let $$r$$ be the real line through $$p,q$$. The plane $$H$$ generated by $$r$$ and a third point $$c\notin r$$ can be parametrized by $$(1-s-t)c+sp+tq$$, and we consider the smooth mapping $$F:(\mathbb R^m\setminus r)\times(\mathbb R^2\setminus\{s+t=1\})\to\mathbb R^m$$ given by $$F(c,s,t)=F_c(s,t)=(1-s-t)c+sp+tq.$$ We exclude $$s+t=1$$ so that all partial derivatives $$\partial F/\partial c=(1-s-t)$$Id with $$1-s-t\ne0$$ are linear isomorphisms, which guarantees that for a residual set $$C_i$$ of $$c$$'s the partial mapping $$F_c$$ is transversal to $$M_i$$ (parametrized version of density of transversality). This implies the inverse image $$F_c^{-1}(M_i)\subset\mathbb R^2\setminus\{s+t=1\}$$ is a manifold of codimension equal to that of $$M_i$$, hence at least $$2$$. Consequently $$Y_i=F_c^{-1}(M_i)$$ has dimension at most $$0$$ and it is a countable set. As $$F_c(Y_i)=M_i\cap H\setminus r$$, we see that $$M_i\cap H\setminus r$$ is countable. In other words, $$H$$ is in general position with $$M_i$$.
Thus $$C=\bigcap_iC_i$$ is residual, hence dense, hence not empty, and any $$c\in C$$ gives a plane $$H$$ transversal to all $$M_i$$'s. Thus all intersections $$M_i\cap H\setminus r$$ are countable, and $$X\cap H\setminus r$$ is countable. We can produce a polygonal path in $$H$$ from $$p$$ to $$q$$ using two lines $$\ell$$ and $$\ell'$$ as above, with the additional precaution that they are chosen different form $$r$$ to avoid the posible intersection $$X\cap r$$.