Gerschgorin circle theorem question How can show the following property of Gerschgorins theorem?

If $N$ of the disk from a connected domain that is disjoint from the other $m-n$ disk then there are $n$ values eigenvalues of $A$ within this domain.

How can I show this?
I know every eigenvalue of A lies in at least one of the $m$ circular disk in the complex plane with center $a_{ii}$ and radius $\sum_{j=i}|a_{ij}|$.
 A: Essentially, that's because eigenvalues vary continuously with matrix entries.
Decompose $A$ as $D+F$, where $D$ is the diagonal part and $F$ is the off-diagonal part. Consider the function $M(t)=D+(1-t)F$ with $t\in[0,1]$. When $t$ goes from $0$ to $1$, the matrix changes from $A$ to $D$, i.e. we shrink the off-diagonal part of the matrix to zero, while leaving the diagonal part unchanged. Put it another way, the centres of the Gersgorin discs thus remain unchanged, but the discs become smaller and smaller and each of them finally becomes a single point.
Clearly, if some union of discs $U$ is disjoint from the union $V$ of the rest at the beginning, the discs in $U$ will remain disjoint from $V$ in the whole course of shrinking. Consequently, any eigenvalue that lies inside $U$ at the beginning must remain inside $U$ the whole time, otherwise it must enter $V$ at some point (because every eigenvalue must belong to at least one Gersgorin disc), but that is impossible because there is no man's land between $U$ and $V$ (as they are disjoint) and the continuous path of eigenvalues cannot enter this no man's land (again, because every eigenvalue at any time must belong to at least one Gersgorin disc).
Similarly, eigenvalues in $V$ must also remain in $V$ as time proceeds. It follows that the number of eigenvalues inside $U$ is invariant throughout the whole journey. Since $U$ contains exactly $|U|$ eigenvalues when all discs are shrunk to single points, it must contain the same number of eigenvalues at the beginning. Hence the assertion follows.
