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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 $r \to 0$?

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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
up vote 5 down vote accepted

The correct generalization would be by speaking of their diameters going to zero. It is not enough that their Lebesgue measure (or, volume, area) goes to zero, as pointed out this can yield other results than singletons. Basicly any "lower dimensional" object has Lebesgue measure of zero, such as a line in the plane.

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 $n\in\mathbb{N}$).

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Nice answer, thank you. – Alex Petzke May 22 '12 at 12:16
Sure, you're welcome. – T. Eskin May 22 '12 at 12:27

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