Singular points of an analytic variety I am struggling with a question that was posed here before, about why a reducible analytic variety $V=V_1\cup V_2$ must be singular in $V_1\cap V_2$. 
I must say I didn't really figure out the suggestions there, but I thought of a different reasoning: basically, a point in $V$ is non-singular if it has a neighborhood homeomorphic to $\mathbb{C}^n$. Now if the intersection is a point $V_1\cap V_2 = \{p\}$, and $p$ is non-singular, then a neighborhood $U$ must have a homeomorphism $\phi(U)=\mathbb{C}^n$. In this case $\phi(V_1-p)\cup\phi(V_2)$ is a disjoint union covering $\mathbb{C}^n$, and so $\mathbb{C}^n$ is not connected.
Does this proof make sense? And can it be generalized for larger $V_1\cap V_2$? (I'm wondering if the intersection must always have a point with a neighborhood that shares points with both sets).
 A: I am also stuck with this question and I am not sure if this will help. Referring to a book "Holomorphic functions and integral Representations in several complex variables" by Michael Range, the two following exercise problems (unfortunately!) can help you answer your question:

[Page 31: E.2.13] Let $M_{1}$ and $M_{2}$ be closed connected complex manifolds of the region $D\subseteq \mathbb{C}^{n}$. If there exists $U$ a neighbourhood of $P\in M_{1}\cap M_{2}$ with $U\cap M_{1}=U\cap M_{2}$, then $M_{1}=M_{2}$.
[Page 40: E.3.8] Let $A_{1}$ and $A_{2}$ be analytic sets, $P\in A_{1}\cap A_{2}$. If for each $U$ neighbourhood of $P$, $U\cap A_{1}\neq U\cap A_{2}$, then $P$ is a singular point of $A=A_1\cup A_2$.

Remark: I think in E.3.8, it is implicitly implied that $A_{1}\neq A_{2}$.
Therefore, if $V_{1}\cap V_{2}$ is not contained in $V_{s}$, then we may find $z\in V_{1}\cap V_{2}$ which is regular. Then by E.3.8, there exists a neighbourhood $U$ of $z$ such that $U\cap V_{1}=U\cap V_{2}$. By $E.2.13$, therefore $V_{1}=V_{2}=V$ which is not possible.
So we are left with finishing two questions. Hope someone can help!
N.B. This is the best I can come up with. It might be wrong also, so exercise caution!
A: *

*$M_1\cap M_2$ is an analytic set of $D$ and therefore it is an analytic set of $M_1$. Since $M_1\supset U\cap M_1=U\cap M_2\subset M_2,$ $U\cap M_1\subset M_1\cap M_2$. So, the analytic set $M_1\cap M_2$ contains a nonempty open subset in  $M_1$, by connectedness, $M_1=M_1\cap M_2$. Do the same to $M_2$.

*If $P$ is not a singular point of $A$, there is a connected neighborhood $U$ of $P$ such that $A\cap U$ is complex submanifold of $U$. Choosing our coordinate system the question reduced to a local case:
The union of two proper varieties of $\mathbb{C}^n$ can not contain a neighborhood of origin.

Suppose that the two varieties at origin are given by zeros of $(f_1,\ldots,f_r)$ and $(g_1,...,g_k)$ respectively. Then the union at the origin is given by the zeros of $(f_i g_j)$. Therefore, by the identical principle, the zeros of $f_ig_j$ contains a neighborhood of origin if and only if $f_i$ is always zero or $g_j$ is always zero. Since the two varieties are proper subsets, this completes the proof.
