# Nested sequences of balls in a Banach space

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 $\bigcap_{n=1}^\infty B_n\neq\varnothing$.

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.

• 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. Feb 17, 2011 at 9:27
• "[...] be a sequence of nested in a banach space". Seems like you missed something there? Should it be "nested closed balls"? Feb 17, 2011 at 9:50
• yeah, it's nested sequence- as the title suggests :) i'm changing it Feb 17, 2011 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. Feb 17, 2011 at 10:05
• @George: Perhaps the simplest example of a complete metric space where this fails is the natural numbers with the metrc $d(m,n)=1+1/(\min (m,n))$. Feb 17, 2011 at 11:35

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 $x_{N} \in \bigcap_{n = 1}^{\infty} B_{\leq r_{n}} (x_{n})$.

• @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, 2011 at 12:19
• 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). Feb 20, 2011 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, 2011 at 4:47
• Did you nowhere use the reflexivity of the Banach space? What about this example? Jul 4, 2013 at 9:16
• Sorry for this comment which is probably stupid but I am wondering how you would prove that $B_r(x) \subset B_s(x)$ implies $r \le s$. What about these examples: math.stackexchange.com/questions/734248/… ? Thanks. Aug 9, 2014 at 10:31