Subanalytic set

Remarks

Every semianalytic set $X \subseteq \R^n$ is subanalytic. Indeed, for a point $a$ take an open ball $U$ around $a$, take $m = 0$ and $A = X \cap U$. This set $A$ is semianalytic and bounded, and $\pi$ is the identity, so $X \cap U = \pi(A)$.

The set $X = \set{1/k : k = 1, 2, 3, \dots} \subseteq \R$ is not subanalytic. A subanalytic set meets each small enough neighbourhood of a point in finitely many connected components. But every neighbourhood of $0$ contains infinitely many points of $X$, each of them a connected component of the intersection.

The same definition works in a real analytic manifold $M$. One replaces $\R^m$ by another real analytic manifold $N$ and asks for a relatively compact semianalytic subset $A$ of $M \times N$.