# What are some subanalytic sets which are not semianalytic?

I'm working with sheaf theory for the most part, and subanalytic sets are an integral part of the field. However, I'm struggling to get an intuition for subanalytic sets. I know, because I've seen it stated several times now, that they are a necessary concept to work with (at least if we're looking to use o-minimality. Otherwise, honestly I still don't quite get the need for them). The argument I've seen being used for their utility is the breakdown of the Tarski-Seidenberg Theorem, meaning that the property isn't preserved under projections essentially. However, I have never seen a proof of this fact, and/or some examples of the differences between semianalytic sets and subanalytic sets.

• Could anyone provide me with some examples/concrete cases that illustrate the difference between the properties, or at least some references that do so? Or even just some papers where it is proven that the Tarski-Seidenberg Theorem fails for semianalytic sets, with preference for the original paper?

Thank you for all the help in advance :)

• Removed analytic-geometry tag, which is not used for anything remotely approaching the level of sheaf theory: analytic geometry -- school level and beginning college level; sheaf theory -- graduate level (and in the U.S., usually beyond both 1st year and Ph.D. Qualifying Exam topics). Feb 12 at 13:12
• Really? Ok, it's such a widely used used expression for geometry on analytic manifolds. Good to know :) Feb 13 at 10:51

So, there is apparently a result by Łojasiewicz (in some lecture notes of his: "Ensembles semi-analytiques", published in 1965) that states the following:

Let $$X \subset M$$ be a subanalytic subset of an analytic manifold. Then, $$X$$ is semianalytic of dimension $$\leq k$$ if and only if locally in $$M$$, there is an analytic set $$Z \subset M$$ of dimension $$\leq k$$ such that $$X \subset Z$$, $$\overline{X} - X$$ is semianalytic of dimension $$\leq k-1$$ and $$X - \text{int}_{Z}(X)$$ (that is, $$X$$ except its interior in $$Z$$) is also semianalytic of dimension $$\leq k-1$$.

Using this criterion, and the concept of dimension attributed to semianalytic sets (using stratification with some more semianalytic sets), an argument by Osgood (in 1916, in an article called "On functions of several complex variables", page 2) can be used, as follows:

Consider the projection \begin{align*} \pi: \mathbb{R}^{4} & \rightarrow \mathbb{R}^{3}\\ (x,y,z,w)&\mapsto (x,y,w), \end{align*} and the semianalytic set (in fact analytic) $$X := \mathcal{Z}(y-xz, w-xe^{z}) =\{ (x,y,z,w) \in \mathbb{R}^{4} \textbf{ } | \textbf{ } y = xz, w = xe^{z} \}$$, i.e. the common zero locus of the two functions (or of the single function $$(x,y,z,w) \mapsto (y-xz)^{2} + (w-xe^{z})^{2}$$ ). Using the Implicit Function Theorem, we see that $$X$$ is locally a 2-dimensional manifold.

Then, if $$\pi(X) = \big \{ (x,y,w) \in \mathbb{R}^{3} \textbf{ } | \textbf{ } \exists{z} \textbf{ } (y = xz \wedge w = xe^{z}) \big \}$$ is semianalytic, by Łojasiewicz, it is locally contained in an analytic set $$\pi (X) \cap U \subset \mathcal{Z}(G(x,y,w))$$, in which, at its smooth points, the dimension can't exceed 2. So, $$G\circ \pi(x,y,z,w) = 0 \textbf{ } \forall{(x,y,z,w) \in X\cap \pi^{-1}(U)}$$

Since $$(x,y,z,w) \in X\cap \pi^{-1}(U) \implies y = xz, \textbf{ } w = xe^{z}$$, the equation above becomes $$G(x,xz,xe^{z}) = 0 \text{ for all }x,z \in \mathbb{R} \text{ such that } (x,xz,z,xe^{z}) \in \pi^{-1}(U)$$

By continuity, this is valid in an open subset of $$\mathbb{R}^{2}$$, say $$V$$.

But, since $$G$$ is analytic, we can decompose it as an infinite sum of homogeneous polynomials (the convergence of the taylor expansion of $$G$$ lets us permute the terms): $$G(x,xz,xe^{z}) = \sum \limits_{i \geq 0} g_{i}(x,xz,xe^{z}) = \sum \limits_{i \geq 0} g_{i}(1,z,e^{z})x^{i} = 0 \text{ (on } V)$$ So, for each $$z \in \pi'(V)$$ ($$\pi': \mathbb{R}^{2} \to \mathbb{R}$$), we necessarily have $$g_{i}(1,z,e^{z}) = 0 \text{, }\forall{i \geq 0}$$

But, since the exponential is a transcendental function, each $$g_{i}$$ must be the zero polynomial and $$G = 0$$. This leads to a contradiction, because then $$\mathcal{Z}(G) = \mathbb{R}^{3}$$ is 3-dimensional, when it had to have dimension at most 2.

So, we conclude that $$\pi(X)$$ is subanalytic but certainly not semianalytic.

We can furthermore use this to get a nice picture :)

Take the proper analytic map $$f: \mathbb{R}^{2} \to \mathbb{R}^{3}$$ given by $$(u,v) \mapsto (u,uv,ue^{v})$$. Then, $$\text{im}(f)$$ is subanalytic (in, fact the image of a subanalytic set under a proper analytic map is still subanalytic, as can be checked in Why is the image of a subanalytic set through a proper analytic map subanalytic?) but it is not semianalytic. This can be checked using the same argument as Osgood did.

Now, a nice picture for $$\text{im}(f)$$ using Mathematica:

Note: the only special thing about the exponential in the expression was that it is transcendental, so we could substitute it for other transcendental functions, and get more nice pictures :)