Let $M$ be a complex manifold and $V\subseteq M$ be an analytic hypersurface i.e. $V$ is locally a zero set of some holomorphic function which can be treated as a defining function of $V$. Want to prove that for each $x\in V$ one can find an open set $U_{x}$ of $M$ containing $x$ s.t. $V\cap U_{x}=\cup_{i=1}^{k}V_{i}\cap U_{x}$, where $V_{i}\cap U_{x}$ is an irreducible component of $V\cap U_{x}$ in $U_{x}$ for each $i=1\ldots k$.
Now, pick an $x\in M$. For any subset $A$ of $M$ define the germ of $A$ at $x$ as the image of $A$ under the map $\mathcal{P}(M)\mapsto \mathcal{P}(M)$/~, where for $X,Y\subseteq M$, $X$~$Y$ if there exists an open set $U$ in $M$ containing $x$ s.t. $X\cap U=Y\cap U$. I'll denote it by $(A,x)$. $(A,x)$ will be called an analytic germ if A is an analytic set in some neighbourhood of $x$ (Analytic sets are locally zero set of finitely many holomorphic functions). Lastly, an analytic germ $(A,x)$ is irreducible if $(A,x)=(A_{1},x)\cup(A_{2},x)$ implies $(A,x)=(A_{i},x)$, $i=1$ or $2$, where $(A_{i},x)$ is an analytic germ for each $i=1,2$.
In the book "Complex Analytic and Differential Geometry" by Jean-Pierre Demailly it has been shown that every analytic germ has a finite decomposition into irreducible analytic germs (P-92, Theorem 4.7). I suspect that this can be used to prove the above. However, I am stuck in the following:
Fix $x\in V$. Then we have $(V,x)=\cup_{i=1}^{k}(V_{i},x)$, where $(V_{i},x)$ is irreducible for each $i=1,\ldots k$. The equality here implies that there exists an open set $U_{x}$ in $M$ containing $x$ s.t. $V\cap U_{x}=\cup_{i=1}^{k}V_{i}\cap U_{x}$. But how irreducibility of $(V_{i},x)$ guarantees an open neighbourhood $W_{x}$ of $x$ s.t. $V_{i}$ is irreducible in $W_{x} ?$ Any help is appreciated.