Let $V\_i$ be a sequence of $k$ dimensional analytic subsets in $\mathbb C^N$. Suppose that the volume of $V\_i$ is uniformly bounded, then Bishop's compactness theorem says that $V\_i$ will convergence by sequence to an analytic subsets $V$.


Q1: What is the precise meaning of "converge" here.
Q2: Is it possible that a sequence of singular point $q\_i\in V\_i$ converge to a smooth point $q\in V$. It seems impossible.



Bishop's compactness theorem and convergence of analytic subset
Convergence is taken [in Hausdorff sense](https://en.wikipedia.org/wiki/Hausdorff_distance),
though you can define the structure of a complex variety (the [Barlet space](http://library.msri.org/books/Book37/files/barlet.pdf))
on the set of cycles, taking every irreducible component with positive integer multiplicity. Barlet convergence is slightly more fine than the Hausdorff convergence, but not by much.


For the second, it is easy to construct an example of singular spaces converging to smooth. Take a sequence of curves in ${\Bbb C}P^2$ with each curve obtained as a union of two projective lines. Assume that this sequence converges to a union of a line with itself. Then the limit is smooth.


