Connectedness and path connectedness, of irreducible affine algebraic set in $\mathbb C^n$, under usual Euclidean topology

Question

Let $n\ge 2$ and $V$ be an irreducible affine algebraic set in $\mathbb C^n$ . Then is it true that $V$ is path connected (or at least connected ) in the usual Euclidean topology of $\mathbb C^n$ ?

Answer 1

Yes, $V$ is connected. The relevant theorem is the following:

Connectedness theorem.
Let $X$ be an irreducible complex projective variety. If $Y \subsetneq X$ is a Zariski-closed strict subset, then $X\setminus Y$ is connected in the classical topology.

Now, if $X=\bar V\subset \mathbb P^n(\mathbb C)$ and $Y=X\cap H$ (where $H=\mathbb P^n(\mathbb C)\setminus \mathbb C^n$=hyperplane at infinity), we have $V=X\setminus Y$ and the theorem applies, proving the connectedness of $V$.

Reference for connectedness theorem: Mumford, Algebraic Geometry I, (4.16) page 68.

Edit: direct proof of path connectedness
Let $p:\tilde V\to V$ be a desingularization of $V$.
For $a,b\in V$ there exist $\tilde a, \tilde b\in \tilde V$ with $p(\tilde a)=a, p(\tilde a)=b$.
Join $\tilde a, \tilde b\in \tilde V$ by a path $\gamma:[0,1]\to \tilde V$ (possible since $\tilde V$ is a connected complex manifold).
The path $p\circ \gamma$ then joins $a$ to $b$, qed.

Note to killjoys :-)
Grumpy people who wish to point out that my proof might not be completely elementary should show Hironaka some respect : he didn't work years on desingularization in order to have spoilsports prohibiting mention of his work!