# Thread: nested interval thm in R^n

1. ## nested interval thm in R^n

how would i prove that the nested interval theorem generalizes in the n-space R^n? the nested interval theorem in the real line R just says that if (I_k:k in N)is a decreasing sequence of bounded closed sets in R (just closed intervals) then we have that the intersection of all I_k's is nonempty. furthermore, if L(I_k) denotes the length of each I_k, and if lim(k->infinite) L(I_k) = 0, then intersection of all I_k's is a singleton.

to prove this theorem in the space R^n would i just consider the L-2 norms of each set?

2. Originally Posted by squarerootof2
how would i prove that the nested interval theorem generalizes in the n-space R^n? the nested interval theorem in the real line R just says that if (I_k:k in N)is a decreasing sequence of bounded closed sets in R (just closed intervals)
A bounded closed set may be something else than just a closed interval or a finite union of closed intervals. It can be pretty complicated as well (cf. Cantor sets).
then we have that the intersection of all I_k's is nonempty. furthermore, if L(I_k) denotes the length of each I_k, and if lim(k->infinite) L(I_k) = 0, then intersection of all I_k's is a singleton.

to prove this theorem in the space R^n would i just consider the L-2 norms of each set?
What is the L-2 norm of a set? You can use the diameter of a set: ${\rm diam}(A)=\sup\{\|x-y\|\,|\,x,y\in A\}$, where $\|\cdot\|$ is any norm.

If you know the 1-dimensional case (not only for intervals), you can prove the general case by considering the projections of the sets on the axes (they are compact as well, i.e. closed and bounded, but this is not a straightforward consequence of being closed and bounded, be careful), and applying the 1-dimensional theorem to these.

Otherwise, it would really help to know that "A is bounded and closed" (in $\mathbb{R}^n$) is equivalent to (or just implies that) "every sequence in A possesses a converging subsequence". (I think this is called Bolzano-Weierstrass property)
If you know that, choose $a_k\in A_k$ for every $k$ (I assume the sequence of sets is $(A_k)_k$). Then there is a converging subsequence (because $\forall k\geq 1,\,a_k\in A_1$) and its limit is in $A_1$, but also in $A_2$ (because $\forall k\geq 2,\, a_k\in A_2$), and so on, so that the limit belongs to the intersection of all the $A_k$: this intersection is non-empty. If the diameters tend to 0, you easily see that two points in the intersection must be arbitrarily close to each other (at a distance less than the diameter of $A_k$, which tends to 0), and hence equal.

3. hmm i am familiar with the bolzano weierstrauss theorem, but i am not allowed to use it in proving this theorem because this theorem is actually what we used to prove the bolzano weierstrauss. by L2 norm i just mean the euclidean norm, but i guess as you suggested the sup norm would work better. thanks!

4. You don't need to consider L^2 norms. Just look at coordinates. If you have a nested sequence of bounded closed sets in R^n then (for j=1,2,...,n) their j'th coordinates form a nested sequence of bounded closed sets in R. By the nested interval theorem for the real line, this sequence has a nonempty intersection. If x_j is a point in this intersection then the point $(x_1,\ldots,x_n)$ is in the intersection of the sets in the R^n sequence.

If you put an extra condition on the sets in R^n, such as asking that their diameters should tend to zero, then their intersection will be a single point.

5. Originally Posted by Opalg
You don't need to consider L^2 norms. Just look at coordinates. If you have a nested sequence of bounded closed sets in R^n then (for j=1,2,...,n) their j'th coordinates form a nested sequence of bounded closed sets in R. By the nested interval theorem for the real line, this sequence has a nonempty intersection. If x_j is a point in this intersection then the point $(x_1,\ldots,x_n)$ is in the intersection of the sets in the R^n sequence.

If you put an extra condition on the sets in R^n, such as asking that their diameters should tend to zero, then their intersection will be a single point.
As such, this projection argument fails, because there no reason why the point $(x_1,\ldots,x_n)$ would lie in the intersection. For sure the $x_i$ are coordinates of some points in the intersection, but no necessarily the same. However, it would suffice to say for instance the projection on the x-axis of the intersection is non-empty, which directly implies that the intersection is non-empty.
However, is the intersection of the projections a subset of the projection of the intersections? Despite what I wrote myself, this seems nontrivial to me. For instance, if we know that for every $k$, $\xi$ is the first component of some $x_k\in A_k$, does it imply that $\xi$ is the first component of an element of the intersection of the $A_k$'s ? This would easily result from the Bolzano-Weierstrass property.

In addition to that, like I said in my post,
- the projections of the sets may not be intervals, so not only the "nested intervals" case is needed;
- the fact that these projections are closed requires a proof, which involves the Bolzano-Weierstrass property (in $\mathbb{R}^n$).

As a conclusion, I really think you can't avoid (except artificially) to first prove the Bolzano-Weierstrass property for bounded closed subsets of $\mathbb{R}^n$: "a sequence in a bounded closed set of $\mathbb{R}^n$ possesses a convergent subsequence". This can be obtained from a similar (or slightly weaker) result on $\mathbb{R}$ saying that bounded sequences have a convergent subsequence. Take a sequence $(x^p)_{p\geq 0}$ in a bounded closed set $K$ of $\mathbb{R}^n$. The sequence $(x^p_1)_{p\geq 1}$ of the first coordinates is bounded, hence possesses a convergent subsequence $(x^{\phi(p)}_1)_{p\geq 0}$; do the same with the second coordinates of the sequence $(x^{\phi(p)})_{p\geq 0}$: there is a converging subsequence $(x^{\phi(\psi(p))}_2)_{p\geq 0}$, and so on. Finally, you have a subsequence whose coordinates all converge, hence it converges in $\mathbb{R}^n$, and the limit belongs to $K$ because this set is closed. But you've probably already seen this proof in your course.

6. Originally Posted by Laurent
Originally Posted by Opalg
You don't need to consider L^2 norms. Just look at coordinates. If you have a nested sequence of bounded closed sets in R^n then (for j=1,2,...,n) their j'th coordinates form a nested sequence of bounded closed sets in R. By the nested interval theorem for the real line, this sequence has a nonempty intersection. If x_j is a point in this intersection then the point $(x_1,\ldots,x_n)$ is in the intersection of the sets in the R^n sequence.

If you put an extra condition on the sets in R^n, such as asking that their diameters should tend to zero, then their intersection will be a single point.
As such, this projection argument fails, because there no reason why the point $(x_1,\ldots,x_n)$ would lie in the intersection.
Oops! That's absolutely correct, of course.

However, I think that my argument can still be made to work if you know that the diameters of the sets tend to zero, essentially because $(x_1,\ldots,x_n)$ is then the only candidate for the point of intersection.