Is a complex space irreducible at a point x also irreducible at any point of a neighborhood of x? Let $X$ be a complex analytic space. For my question it suffices to assume that $X$ is reduced.  Let $x\in X$ and assume that $X$ is irreducible at $x$, which means that the stalk $\mathscr O_{X,x}$ is an integral domain. Is it true that there is a neighborhood $U$ of $x$ such that $X$ is irreducible at every point of $U$?
 A: No, in the book Complex Analytic Sets written by E. M. Chirka, page 56, Example (c), the author provided the following example meaning that the set of irreducibility points of an analytic set $A$ need not be open or closed in A.
Example: the singular points of the analytic set $A:z_1^2=z_3z_2^2$ in $\mathbb C^3$ form the $\mathbb C^3$-axis, whild $\textit{reg }A$ is the graph $z_3={\left(\frac{z_1}{z_2}\right)}^2$ above $\mathbb C^2\setminus \left\{z_2=0\right\}$. So, $\textit{reg }A\cap V $ is connected in any polydisk neighborhood $0\in V$, hence $A$ is locally irreducible at $0$. However, it is clear that $A\cap V$ is reducible in any domain $V$ contains $\left(0,0,z_3\right)$, here $z_3\not=0$, and we should ensure the diameter of $V$ does not exceed $|z_3|$. In summary, there exist reducible points of the analytic set converging to
the origin.

In fact, this example  was first proposed by Osgood in his article
W. F. Osgood, Lehrbuch der Funktionentheorie, II. 1, 2nd edition, Leipzig, 1929.
