---
tag: F55M92H5JY
kind: definition
name: Subanalytic set
msc: [32B20, 14P15]
defines: K1HTQF26MA
uses: [E1V22QQHYJ, 033DA1WE8Y]
sources:
  - url: https://en.wikipedia.org/wiki/Subanalytic_set
---

## Definition

Let $n \ge 1$ and let $X \subseteq \R^n$.
The set $X$ is *subanalytic* if each point $a \in \R^n$ has an open neighbourhood $U$ with the following property.
There are an integer $m \ge 0$ and a bounded [[semianalytic set|E1V22QQHYJ]] $A \subseteq \R^{n+m}$ such that $X \cap U = \pi(A)$.
Here $\pi \colon \R^{n+m} \to \R^n$ is the [[projection|033DA1WE8Y]] onto the first $n$ coordinates.

## 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 $M \times N$.