Consider a $\textbf{flat}$ surjective map $f: X \rightarrow \mathbb{A}^1$. The general fibers $F\_{\epsilon}$ are canonically isomorphic, and the special fiber $F\_0$ above $0 \in \mathbb{A}^1$ is not isomorphic to the general fibers. 


Given a closed subscheme $B \subset X$, we define its special fiber limit $\widetilde{B}$ to be the intersection of the special fiber $F\_0$ with the closure of $B \times \mathbb{A}^1 \backslash \{ 0 \}$ in $X$, 
$F\_0 \cap \overline{B \times \mathbb{A}^1 \backslash \{ 0 \}}$.


Let $B\_1$ and $B\_2$ be two closed subschemes of $X$. When does the limit of their intersection equal to the intersection of their limits, i.e. 
$\widetilde{B\_1} \cap \widetilde{B\_2} = \widetilde{B\_1 \cap B\_2}$? Is it enough for the dimension of $\widetilde{B\_1} \cap \widetilde{B\_2}$ to be equal to the dimension of $B\_1 \cap B\_2$? Any relevant comments and references are welcome. Thanks!



Flat family: limit of intersection vs intersection of limits
Obviously $\widetilde{B\_1} \cap \widetilde{B\_2} \subseteq \widetilde{B\_1\cap B\_2}$. I'll discuss a sufficient condition for the reverse.


If $B\_1\cap B\_2$ is equidimensional, then so is $\widetilde{B\_1\cap B\_2}$. It seems like you know that its dimension is that of $\widetilde{B\_1}\cap \widetilde{B\_2}$. By any chance is $\widetilde{B\_1}\cap \widetilde{B\_2}$ reduced, and the family projective? Then if $\deg(B\_1\cap B\_2) = \deg(\widetilde{B\_1}\cap \widetilde{B\_2})$ (w.r.t. the projective embedding), your $\widetilde{B\_1\cap B\_2}$ has to be all of $\widetilde{B\_1}\cap \widetilde{B\_2}$.


