Path connectedness equivalent to connected in euclidean topology for quasiprojective vareities?

Question

Let $V$ be a quasiprojective real variety. My intuition tells me that this type of space with the Euclidean topology has the property that path-connectedness and connectedness are equivalent. Is this true? If so, is there an easy argument for this?

I would expect the result does not change if I replace real with complex or quasiprojecive variety with semi-algebraic set.

Answer 1

Yes! One easy argument is that any such variety admits a triangulation, and any set admitting a triangulation is locally path connected which implies that path connected is equivalent to connected.

The first claim about triangulations can be sourced to Hironaka. Here is his original presentation, which is a fair bit more general than you will need:

Theorem (Hironaka, Triangulations of Algebraic Sets in Algebraic Geometry, Arcata 1974 p.180): Given a locally finite system of subanalytic sets $\{X_\alpha\}$ in $\Bbb R^n$, there exists a simplicial decomposition $\Bbb R^n=\bigcup_a \Delta_a$ and a subanalytic automorphism $\kappa$ of $\Bbb R^n$ such that

1. each $X_\alpha$ is a locally finite union of some of the $\kappa(\Delta_a)$ and
2. $\kappa(\Delta_a)$ is a locally closed smooth real-analytic submanifold of $\Bbb R^n$ and $\kappa$ induces a real-analytic isomorphism $\Delta_a\stackrel{\sim}{\to}\kappa(\Delta_a)$ for every $a$.

In fact, a version of this claim is true more generally in any o-minimal geometry:

Theorem (van den Dries, Tame topology and O-minimal structures, p. 130): Let $S\subset\Bbb R^m$ be a definable set, with definable subsets $S_1,\cdots,S_k$. Then $S$ has a triangulation in $R^m$ that is compatible with these subsets.

Morally, any sort of algebraic structure is too well-behaved for anything "topologically awful" to happen.

Answer 2

You already mentioned Real Algebraic Geometry by Bochnak, Coste and Roy. An even stronger result is proven there:

Proposition 2.5.13 (p. 42) : A semi-algebraic set is semi-algebraically connected iff it is semi-algebraically path connected

For semi-algebraic sets semi-algebraically connected is however equivalent to being connected (see for instance Theorem 2.4.5 of the same book). I believe, these results were originally proven by Łojasiewicz and Thom. In the comments Hironaka was cited. I think Hironaka gave the credit to Łojasiewicz (I just browsed through it).