Example for $A,B$ non-empty, bounded, and closed, $A\cap B = \emptyset$, but $\operatorname{dist}(A,B) = 0$

Question

Is there such an example in a Banach space? Note that this is not possible in a finite-dimensional space.

In what I have here, $A$ and $B$ are the closures of relatively open sets in a bounded and closed set.

Some suggestions: Using Bolzano-Weierstrass property we can show that in this case A,B must be bounded sets that are not totally bounded. I suggest look into examples like the unit ball in an infinite dimensional Banach space. — Henricus V., Oct 14 '17 at 02:00
@Henricus Of course, non of them can be totally bounded. Otherwise the same argument as in the finite-dim. situation can be applied. — amsmath, Oct 14 '17 at 02:04

Henricus V. · Accepted Answer · 2017-10-14T14:25:41.963

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Inspired by this:

Consider the Banach space $\ell^2$. Define $A$ to be the set of sequences for which only one term is $1$. This is bounded and non-empty and closed.

Define $B$ to be the set of sequences of the form $$ \mathbf b = (0,\dots,0,1+2^{-n},0,\dots) $$ where $n$ is the index of that non-zero term in the sequence. This is also bounded (all terms are within 2 distance from 0) and non-empty and closed.

$A,B$ are disjoint.

Then for all $\delta > 0$ we can find $\mathbf a \in A$ and $\mathbf b \in B$ with $\lVert \mathbf a - \mathbf b\rVert < \delta$. Indeed, take $\mathbf a$ to be the sequence for which $a_n = 1$ and $\mathbf b$ to be the sequence for which $b_n = 1 +2^{-n}$ for the same $n$. For sufficiently large $n$ $1 + 2^{-n} - 1$ can be arbitrarily small. Hence $\operatorname{dist}(A,B) = 0$.

edited Oct 14 '17 at 14:25

answered Oct 14 '17 at 02:06

Henricus V.

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Thank you very much, dear Henricus. :o) – amsmath Oct 14 '17 at 02:13
@amsmath I learned something new today too. My intuition always told me this example is impossible :) – Henricus V. Oct 14 '17 at 02:14
Same here. I am quite surprised. – amsmath Oct 14 '17 at 02:16

Eclipse Sun · Answer 2 · 2017-10-14T03:57:59.140

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Note that the distance is $0$ iff $0$ is in $\overline{A-B}$.

But it is known that the sum of two closed subsets need not to be closed. See this. (In the link, we find two subspaces $U,V$ of $X$ such that $U+V\ne X$ but $\overline{U+V}=X$. To get a bounded example, we take unit balls $A,B$ of $U,V$. Then $A+B$ is not closed. This is obvious in that example, but I am not sure if this holds in general.)

Hence, you can take such $A,B$ and some point $x\in \overline{A-B}\setminus A-B$. You can assume $x=0$. This will provide an example you want.

edited Oct 14 '17 at 03:57

answered Oct 14 '17 at 02:06

Eclipse Sun

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I cannot follow your argumentation. Also, you seem to wanna show that such an example does not exist (which is not true, considering Henricus' answer). – amsmath Oct 14 '17 at 02:15
I add some explanation. – Eclipse Sun Oct 14 '17 at 02:23
Why can I assume $x = 0$? – amsmath Oct 14 '17 at 02:36
Otherwise you take $A=A-{x}$ instead. – Eclipse Sun Oct 14 '17 at 02:42
Translation of closed sets are closed. – Eclipse Sun Oct 14 '17 at 02:45
Ok. The only problem that's left for me here is that the sets in your link are not bounded... – amsmath Oct 14 '17 at 02:47
Oh I see. You can take closed unit balls instead of subspaces in that link. I will modify my answer. – Eclipse Sun Oct 14 '17 at 03:40
I tried to show that if $M\cap B_1 + N\cap B_1$ is closed, then so is $M+N$, but I couldn't succeed. (Here, $B_1$ is the closed unit ball) – amsmath Oct 14 '17 at 03:44

Example for $A,B$ non-empty, bounded, and closed, $A\cap B = \emptyset$, but $\operatorname{dist}(A,B) = 0$

2 Answers2