Recall that a structure on $\Bbb R$ is defined to be a sequence $\mathcal S=(S_n)_{n<\omega}$ such that for each $n\geq 0$ the following properties hold:
- $S_n$ is a boolean algebra of subsets of $\Bbb R^n$.
- If $A\in S_n$, then $\Bbb R\times A$ and $A\times\Bbb R$ belong to $S_{n+1}$.
- $\{(x_1,\ldots,x_n)\in\Bbb R^n\mid x_1=x_n\}\in S_n$
- If $A\in S_{n+1}$, then $\pi(A)\in S_n$, where $\pi\colon\Bbb R^{n+1}\to\Bbb R^n$ is the projection map on the first $n$ coordinates.
Moreover a structure $\mathcal S$ on $\Bbb R$ is called $o$-minimal if it additionally satisfies:
- $\{(x,y)\in\Bbb R^2\mid x<y\}\in S_2$.
- The sets in $S_1$ are finite unions of intervals (including degenerate intervals, that is singletons).
Recall also that a set $X\subseteq\Bbb R^n$ is called definable with respect to a structure $\mathcal S$ on $\Bbb R$ if $X\in S_n$, and that a function $f\colon\Bbb R^n\to\Bbb R^m$ is called definable if its graph is definable in $\Bbb R^{n+m}$.
Let $X\subseteq\Bbb C^n$ be a closed analytic subset, meaning that $X$ is closed in $\Bbb C^n$ and for every $x\in X$ there is an open subset $U_x\subseteq\Bbb C^n$ and a finite family of holomorphic functions $f_1,\ldots,f_k\colon U_x\to\Bbb C$ such that $X\cap U_x=V(f_1,\ldots,f_k)$. Assume moreover that $X$ is definable is some $o$-minimal structure over $\Bbb R$ (identifying $\Bbb C^n$ with $\Bbb R^{2n}$).
We say that $X$ is irreducible if $X$ cannot be written as the (not necessarily disjoint) union of two proper closed analytic subsets.
Is it true that the irreducible components of $X$ are also definable in this $o$-minimal structure? I only know that this is the case for the connected components of $X$, but of course there are connected reducible closed analytic subsets of $\Bbb C^n$. I thought it would be possible to show this through a cell decomposition of $X$, but I can't see a way to make this kind of argument work.