Embeddings and triangulations of real analytic varieties

This is a follow up question to my answer here How do you define the Euler Characteristic of a scheme?

A real analytic space is a ringed space locally isomorphic to $(X,O/I)$ where $X$ is the zero locus of some number of real analytic functions $f_1,\ldots, f_k$ on an open set $U$ of $\mathbf{R}^n$, $O$ is the sheaf of germs of real analytic functions on $U$ and $I$ is the ideal sheaf generated by $f_1,\ldots, f_k$ (see e.g. http://eom.springer.de/a/a012430.htm) I would like to ask if it is true that each real analytic space with a countable base can be embedded as a closed analytic subset of some Euclidean space.

The motivation behind this comes from the triangulation theorem for complex algebraic varieties: the only proof of that that I know of (Hironaka's 1974 notes) is based on triangulating analytic subvarieties of Euclidean spaces. So to apply this one must embed a complex algebraic variety as a real subvariety of a Euclidean space. This is easy for projective varieties and is probably possible in general, but I don't know a reference for the general case. (I'm mainly interested in the complex algebraic case, but I don't see why it should be any easier that embedding arbitrary real analytic spaces; however if it is easier, I'd be interested to know.)

A related question: is it possible to prove the triangulation theorem (for complex algebraic varieties or in general) without using embeddings in Euclidean spaces?

• I imagine Hironoka's triangulation theorem is a mild tweak of Whitehead's theorem that smooth manifolds admit triangulations. The idea of the proof is to take any embedding of the manifold $M$ into Euclidean space, subdivide a triangulation of Euclidean space sufficiently and keep $M$ transverse to the strata of the triangulation, then pull back the stratifications back to $M$, giving a smooth polyhedral decompostion of $M$. Subdivide to a triangulation of $M$. So you really only need an embedding of $M$ in some object that admits a triangulation, and enough flexibility to get transversality. – Ryan Budney Aug 19 '10 at 6:09
• Are you asking if a real analytic space is a real affine algebraic variety? What kind of embedding are you happy with? There's a characterisation of closed subsets of Euclidean space as something like all spaces that have finite Lebesgue covering dimension, are Hausdorff and are 2nd countable. Perhaps I'm forgetting a criterion, but it's something that appears in many point-set topology texts. – Ryan Budney Aug 19 '10 at 6:30
• Ryan -- Hironaka's proof (based on an earlier proof by Lojasiewicz) is by projecting and using induction on the dimension. The idea you mention probably works as well, but some modifications will be necessary: eg the fact that a triangulation is transverse to each stratum does not guarantee a polyhedral decomposition: take a simplex in the 3-space that contains the vertex of a quadratic cone and intersects transversally each stratum. In general the local structure of real analytic spaces can be pretty messy. – algori Aug 19 '10 at 12:44
• Re what kind of embedding I'm looking for: as a closed analytic subset. Will clarify that in the posting. I'm not sure point set topology does the trick since it is about continuous embeddings and the image can be a complete mess to which the triangulation theorem does not apply. – algori Aug 19 '10 at 12:50
• @algori:There is a proof of triangulation of real analytic spaces by B Giesecke Math Zeutschrift vol 83 ,pages 177-213 yr 1964 – Mohan Ramachandran Jun 13 '11 at 16:08