Cantor's Nested Intervals Theorem can be stated as "If $\{[a_n,b_n]\}_{n=1}^\infty$ is a nested sequence of closed and bounded intervals, then $\cap_{n=1}^\infty [a_n,b_n]$ is not empty. If, in addition, the diameters of the intervals converge to $0$, then $\cap_{n=1}^\infty [a_n,b_n]$ has precisely one member."
How exactly does this generalize to $\mathbb{R}^2$? The first part I think is pretty straightforward: 'If $\{[a_n,b_n] \times [c_n,d_n]\}_{n=1}^\infty$ is a nested sequence of closed and bounded rectangles in $\mathbb{R}^2$, then $\cap_{n=1}^\infty[a_n,b_n] \times [c_n,d_n]$ is nonempty.'
It's the second part that I'm curious about. The natural two-dimensional analogue of the diameter of the intervals converging to $0$ would be the area of the rectangles going to zero. However, this could result in a point or a line segment. If $a$,$c$ and $b$,$d$ are the supremums and infimums respectively of the sequences of endpoints of the intervals $[a_n,b_n]$ and $[c_n,d_n]$, is the correct generalization that the intersection will be $[a,b] \times \{c=d\}$, $\{a=b\} \times [c,d]$, or a single point, $\{(a,c)\}$? Or is it more correct to say that if $\mbox{diam}[a_n,b_n] \to 0$ and $\mbox{diam}[c_n,d_n] \to 0$ then the intersection is a single point, $\{(a,c)\}$. The latter doesn't require us to make 'area' meaningful so I have the feeling this is the case but I'd like to hear some thoughts on it.
Afterthought: Does a similar idea work for nested sequences of closed balls $B[x,r]$ as $r \to 0$?