TITLE: Is the singular homology of a real algebraic set always finitely generated?
QUESTION [13 upvotes]: Here is a precise statement of my question:
Let $p\in \mathbb{R}[x_1, \ldots, x_n]$ be a polynomial, and let $Z(p)\subset \mathbb{R}^n$ be the set of zeros of $p$.  Must the singular homology $H_i (Z(p); \mathbb{Z})$ ($i\geq 0$) be finitely generated as an abelian group?  
Here I really just mean the singular homology groups of this set as a topological space with the Euclidean topology.
It's an old theorem of Whitney that $Z(p)$ has finitely many connected components, so $H_0 (Z(p); \mathbb{Z})$ is finitely generated.  Note that it is possible to triangulate $\mathbb{R}^n$ with $Z(p)$ as a subcomplex, so $H_i (Z(p)) = 0$ for $i>n$.
I'm guessing the answer is well-known (either a theorem or a counterexample), but I couldn't find an answer on Google or MathSciNet...

REPLY [10 votes]: Proposition 1.13 of these notes by Coste implies (if I've read it correctly) that any semialgebraic set $S$ is homeomorphic to a union $U$ of open simplices in some finite simplicial complex $K$.
Thus $S$ is an open subset of an ENR, hence is an ENR. Its singular cohomology therefore coincides with its Čech cohomology, which is finitely generated. Therefore its singular homology must be finitely generated.
Edit: The argument given above is incomplete, since a union of "open" simplices in a simplicial complex is not necessarily open (an easy mistake to make!). However it is easy to see that $S\subseteq\mathbb{R}^n$ is an ENR by other means (in particular, it is locally compact and locally contractible).