Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This seems to be a fairly easy question but I'm looking for new points of view on it and was wondering if anyone might be able to help

(by the way- this question does come from home-work, but I've already solved and handed it, and I'm posting this out of interest, so no HW tag)

Let $B_n=B(x_n,r_n)$ be a sequence of nested closed balls in a Banach space $X$ prove, that $\bigcap_1^\infty B_n\neq\emptyset$

As I said before, it should be rather simple. When the radii decrease to 0, it's just a matter of selecting any sequence of points in $B_n$, and it must be Cauchy- and the limit is in the intersection.

My question is what to do when the radii do not decrease to 0? I got some tips about multiplying the balls by a sequence of decreasing scalars, or reducing the radii so that they decrease to 0, but found too many pathological cases for both methods.

Finally- I used a geometric arguemnt (which i've shown to work in any normed space) that if $B(x_1,r_1)\subset B(x_2,r_2)$ then $\| x_1-x_2\|\leq|r_1-r_2|$. This turned out to be some kind of technical catastrophe, but it worked...

Still, if anyone knows of a more elegant solution, I'd love to hear about it

Thanks

share|improve this question
    
What's so catastrophic about the geometric argument? All you need is that the affine line spanned by $x_1$ and $x_2$ is isometrically isomorphic to the ground field, then it's just a 1d (or 2d if your space is complex) picture. –  Chris Eagle Feb 17 '11 at 9:27
    
"[...] be a sequence of nested in a banach space". Seems like you missed something there? Should it be "nested closed balls"? –  kahen Feb 17 '11 at 9:50
    
yeah, it's nested sequence- as the title suggests :) i'm changing it –  kneidell Feb 17 '11 at 9:56
    
It's not true in general. That property is called spherical completeness and fails for the p-adic complex numbers, for example. –  George Lowther Feb 17 '11 at 10:05
1  
@George: Perhaps the simplest example of a complete metric space where this fails is the natural numbers with the metrc $d(m,n)=1+1/(\min (m,n))$. –  Chris Eagle Feb 17 '11 at 11:35
show 1 more comment

1 Answer

up vote 11 down vote accepted

I don't know if this is more elegant, but that's about the best I can come up with at the moment and probably essentially the same as your argument.


Consider first the situation $B_{\leq r}(x) \subset B_{\leq s}(y)$. It is easy to see that $r \leq s$.

Claim. $\|y - x\| \leq s - r$.

Proof. If $x = y$ there is nothing to prove, so let's assume $x \neq y$. The point $z = x - r \frac{y-x}{\|y - x\|}$ belongs to $B_{\leq r}(x)$ and hence also to $B_{\leq s}(y)$. Therefore $\|y - z\| \leq s$. On the other hand, \[ y - z = y - x + \frac{r}{\|y - x\|} (y - x) = \underbrace{\left(1 + \frac{r}{\|y - x\|}\right)}_{\lambda} (y - x), \] so $s \geq \lambda \|y - x\| = \|y - x\| + r$ and hence $\|y - x\| \leq s - r$.


This means that a nested sequence of closed balls $B_{\leq r_{n}}(x_{n})$ has the following properties:

  1. The sequence $r_{n}$ is monotonically decreasing, hence converges to some $r$.
  2. If $N$ is such that $r_{N} \leq r + \varepsilon$ then the above claim implies that for all $n\geq m \geq N$ we have $r_m - r_n \leq \varepsilon$, so $\|x_{m} - x_{n}\| \leq \varepsilon$ because $B_{\leq r_{n}}(x_{n}) \subset B_{\leq r_{m}}(x_{m})$.

In other words, the centers $x_{n}$ form a Cauchy sequence and their limit point $x$ must belong to $\bigcap_{n = 1}^{\infty} B_{\leq r_{n}}(x_{n})$.


Added: As Jonas pointed out, the argument can be made even simpler and doesn't need completeness: Suppose $r_{n} \to r \gt 0$. Then there is $N$ such that $r_{N} \leq 2r$. Then for all $n \geq N$ we have $r \leq r_{n} \leq r_{N} \leq 2r$, so $r_{N} - r_{n} \leq r$ and the claim implies that $\|x_{n} - x_{N}\| \leq r \leq r_{n}$, so $x_{N} \in \bigcap_{n = 1}^{\infty} B_{\leq r_{n}} (x_{n})$.

share|improve this answer
    
@kneidell: While writing this I have forgotten about the fact that you've already had this solution. I don't think there is a much easier way. –  t.b. Feb 17 '11 at 12:19
4  
If the limit $r$ of the radii is positive, then $r_N<2r$ for some $N$, and then $x_N$ will be in every subsequent ball (by the claim). –  Jonas Meyer Feb 20 '11 at 4:33
    
@Jonas: Right. So you needn't even use completeness in that case. Nice! This hasn't occurred to me before, thanks! –  t.b. Feb 20 '11 at 4:47
    
Did you nowhere use the reflexivity of the Banach space? What about this example? –  Student Jul 4 '13 at 9:16
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.