# 'Nested Intervals Theorem' in $\mathbb{R}^2$

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 $r \to 0$?

-
One could ask for the perimeter of the rectangle to go to zero. –  Gerry Myerson May 22 '12 at 2:53
@AlexPetzke I would say that the two-dimensional analogue of the diameter of an interval is the diameter of a circle. –  azarel May 22 '12 at 2:55
One could say that the rectangles or converge to a set of lower dimension. –  Neal May 22 '12 at 3:02
You can conclude that the intersection is a line segment (with points being a special case of a line segment of length 0). –  Arturo Magidin May 22 '12 at 3:17
@azarel: I didn't share this, but the book I'm using sets it up with rectangles, hence viewing it that way. –  Alex Petzke May 22 '12 at 12:06

In fact, the second part of the theorem generalizes to any complete metric space $(X,d)$ by considering a sequence $(F_{i})_{i=1}^{\infty}$ of nested non-empty closed sets such that diam$(F_{i})\to 0$. Here's a sketch how to prove it. By choosing a sequence with $x_{i}\in F_{i}$ for all $i$ we obtain a Cauchy sequence (since the diameters go to zero) which has a limit $x\in F_{1}$ since $X$ is complete and $F_{1}$ is closed. Now it is not hard to show that $x\in \cap_{i=1}^{\infty}F_{i}$ since if there was an index $i_{0}$ with $x\notin F_{i_{0}}$, then $x\notin F_{n}$ for all $n\geq i_{0}$ (since the sequence of sets is nonincreasing). Since $F_{i_{0}}$ is closed its complement is open, so there exists $r>0$ so that $B(x,r)\subset F_{i_{0}}^{c}$. On the other hand, we find $k_{0}$ (by definition of convergence) so that $d(x,x_{i})<r$ for all $i\geq k_{0}$. Choose $n_{0}=\max\{k_{0},i_{0}\}$, whence $d(x,x_{i})$ is less than $r$ (by convergence) and more or equal than $r$ (since the sequence continues in the complement of $B(x,r)$) for all $i\geq n_{0}$, which is a contradiction. Hence $x\in \cap_{i=1}^{\infty}F_{i}$. If there would be more elements in the intersection, say another point $y$, we find $F_{i}$ with diameter less than their distance (since the diameters go to zero) and yet containing both points, which is again a contradiction. Hence the intersection contains only $x$, i.e. it is a singleton.
So to your last question: Yes it works for a nested sequence of closed balls with diameters going to zero by the result above if the underlying metric space is complete (and $\mathbb{R}^{n}$ is complete for all $n\in\mathbb{N}$).