3

I apologize for the stupid question, but I am getting a bit crazy about this.

Consider a Banach space $X$ and a sequence of nested closed balls $(B_n)_n$, i.e. $B_{n+1} \subset B_n$. Let $r_n$ be the radius of $B_n$, for every $n \in \mathbb N$. Question: does the limit $$ \lim_n r_n $$ always exist?

In principle, I would say: "Yes, of course: the sequence $r_n$ must be monotonically decreasing, hence the limit exists. In particular, if it is $0$ then the intersection $\bigcap_ n B_n$ is a singleton while, if $r_n \to r$, then $\bigcap_ n B_n$ is a closed ball of radius $r$."

I am now worried about monotonicity: consider, for instance, these examples. Something related also here and here (note that in the last link the existence of the limit is the hypothesis). Also the post of t.b. here (in particular the first lines) may be relevant.

Thanks.

Community
  • 1
Romeo
  • 6,087
  • Warning: an intersection of balls of decreasing radii $r_n\to r$ need not be a ball; in fact, it may be empty! – Andrew Dudzik Aug 09 '14 at 12:32
  • @you-sir-33433 Sure, thanks for pointing it out; but if you are in a Banach space and the balls are nested then necessarily the intersection is not empty. – Romeo Aug 09 '14 at 13:51
  • No, that's not the case. Let $X=\mathbb{Q}_p$, the field of $p$-adic numbers. Since there is a countable dense subset, $\mathbb{Q}\subset X$, if we pick a nested sequence of balls that avoids every element of $\mathbb{Q}$, and with radii converging to a positive number, then the intersection must be empty. – Andrew Dudzik Aug 09 '14 at 14:08
  • (though you may be thinking specifically of real Banach spaces, in which case you are correct) – Andrew Dudzik Aug 09 '14 at 14:10
  • That's interesting and surprising for me, many thanks. Can I ask you which is the problem for $\mathbb Q_p$, thus? Why does Henning Makholm's proof below does not apply in thi case? I am not very familiar with $p$-adic numbers, to be honest. Thanks again for your comments. – Romeo Aug 09 '14 at 15:06
  • What Henning Makholm's said still applies. The issue with the p-adics (and, generally, ultrametric spaces) is that it is impossible for two distinct balls of the same radius to intersect. Completeness implies that a nested sequence of balls with radii converging to zero must have exactly one point in the intersection. You would think that this implies the same result for radii not converging to zero—just shrink the balls—but this doesn't work in the p-adics, for the reason I mentioned. – Andrew Dudzik Aug 09 '14 at 17:57

1 Answers1

2

Except in the trivial case of the zero space, being Banach prevents pathological examples such as you link to.

More precisely, in a Banach space with at least one nonzero vector, we cannot have $B(x,R)\subseteq B(y,r)$ with $R>r$.

You can see this by considering the two balls restricted to a line that contains $x$ and $y$, with the induced metric. (In the case $x=y$ you need to assume that the space is not the zero space in order to choose such a line). The line is isometric to $\mathbb R$, and the balls intersect it at balls of the same radii on the line. And certainly on the real line, a larger ball cannot be contained in a smaller one.

hmakholm left over Monica
  • 286,031