Intersection of decreasing sequence of non empty closed sets

Question

Let X is Banach space. Suppose $B_n$ are decreasing sequence of non empty closed balls. Prove their intersection is non-empty.

I have some idea. Idea is pick $x_n \in B_n\backslash B_{n-1}$ in such way that $x_n$ is Cauchy sequence. Then $x_n$ will be Cauchy in each ball and each ball is closed so complete hence Cauchy sequence must converge in each ball and limit point is in each ball so it is non-empty.

Main think I don't know how to construct sequence so that it Cauchy.

Anyhelp Please!!!

Compare An empty intersection of decreasing sequence of closed sets. — Martin R, Nov 08 '15 at 19:47
@MartinR A Banach space unequal to ${0}$ is never compact... — Henno Brandsma, Nov 08 '15 at 19:49

score 2 · Accepted Answer · edited Apr 13 '17 at 12:20

2

False, even in the reals. Try $B_n = [n, \infty)$ there.

It is true if $\operatorname{diam}(B_n) \rightarrow 0$ as $ n \rightarrow \infty$, but then this is just Cantor's intersection theorem, which holds in all complete metric spaces (not just Banach spaces).

In the latter case, just picking $x_n \in B_n$ already guarantees Cauchyness of $(x_n)_n$, and then the limit has to be proved to be in the intersection.

It seems to be true for closed balls, see this question , and might be the intended interpretation.

edited Apr 13 '17 at 12:20

Community

1

answered Nov 08 '15 at 19:45

Henno Brandsma

242,131

it does not say diam$(B_n)$ goes to zero but it does say $B_1$ contains $B_2$ contains $B_3$ and so on. – Toeplitz Nov 08 '15 at 19:53
@Toeplitz that is not enough, as my first example shows. – Henno Brandsma Nov 08 '15 at 19:53
@Henno.what if all $B_n$ are closed ball? – Toeplitz Nov 08 '15 at 20:21
@Toeplitz see edit – Henno Brandsma Nov 08 '15 at 20:36

Intersection of decreasing sequence of non empty closed sets

1 Answers1