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Suppose, $X$ is a banach space. For any $x,y \in X$, we define $d(x,y) = |x-y|$, For any $A, B \subseteq X$, we define:

$$d(A, B) = \inf_{x \in A, y\in B}{d(x,y)}$$

Say,$(K_n)_{n \in \mathbb{N}}$ is a Cauchy sequence of disjoint and closed subsets of$X$. We say that:

$$K = \{x \in X: \lim_{n \to \infty}d(\{x\}, K_n) = 0\}$$ is the limit of $(K_n)_{n \in \mathbb{N}}$.

Is $K$ necessarily non-empty?

Added: Thank you all for your examples and corrections. My naive attempt is to find a condition stronger than sequential completeness. I hope it will work by adding "disjoint and closed".

Community
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Metta World Peace
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  • Sorry, but what is a Cauchy sequence of subsets? – Tomás Feb 28 '13 at 18:18
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    What metric are you using on the space of sets? The function above is not a distance. $d([0,1],[1,2]) = 0$. – copper.hat Feb 28 '13 at 18:19
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    SO: consider the three sides of a triangle. They all have distance zero from each other, so if you make a sequence repeating them cyclically it is Cauchy in a very strong sense. But the limit is empty. – GEdgar Feb 28 '13 at 18:25
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    This $d$ does not satisfy separation, as pointed out by copper.hat, but it satsifies symmetry and triangular inequality. It's a pseudometric. – Julien Feb 28 '13 at 18:33
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    Let $K_n = (n,\infty)$. $d(K_n,K_m) = 0$, but $K = \emptyset$. – copper.hat Feb 28 '13 at 18:34
  • @GEdgar Great example. You should make it an answer. – Julien Feb 28 '13 at 18:35
  • @copper.hat Great example too. Same remark. – Julien Feb 28 '13 at 18:36
  • @MettaWorldPeace: You have also $d([0,1],(1,2])=0$. – Seirios Feb 28 '13 at 18:38
  • @copper.hat: Thank you for your illuminating comment. I hope "closed and disjoint" will work. – Metta World Peace Feb 28 '13 at 18:46
  • @Seirios: My bad, thank you. – Metta World Peace Feb 28 '13 at 18:46
  • @Tomás: Sorry for not formulating this problem correctly initially. I hope this version will work. – Metta World Peace Feb 28 '13 at 18:55
  • Notice that if $K \ne \emptyset $ then We can find $N_1$ such that $d(K_n,{x}) < 1/n$ for all $n \ge N_1$ and thus we can also find a $x_1\in K_{N_1}$ such that $d(x_1,x) < 1/n$ continuing like this we get a sequence ${x_i}_{i\in \mathbb{N} ; i \ge 1} $ which converges to $x$ hence we see that if $K$ is non-empty there exists a sequence from $\cup K_i$ that intersects each $K_i$ at most at only one point, converging to some element in $K$ and it is easy to see that if we can create a convergent sequence from $\cup K_i$ that intersects each $K_i$ in one point or less then $K$ is none empty – Oliver E. Anderson Feb 28 '13 at 19:47
  • Maybe we can use the Cauchy property and the fact that the sets are closed to create such a sequence? – Oliver E. Anderson Feb 28 '13 at 19:49

2 Answers2

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Taking my comment, adjusting to make it closed and disjoint. In the plane, let $A_0, A_1, A_2$ be the three sides of an equilateral triangle, all of length $1$. They are closed but not disjoint. Define sequence $K_n$ as follows: If $n=3m$, let $K_n$ be $A_0$ moved radially outward by distance $1/n$. If $n=3m+1$, let $K_n$ be $A_1$ moved radially outward by distance $1/n$. If $n=3m+2$, let $K_n$ be $A_2$ moved radially outward by distance $1/n$. Then this is a Cauchy sequence, $d(K_i,K_j) \le 1/i+1/j$, but its "limit" is empty.

GEdgar
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Let $\Omega \subset \mathbb{R}$ be a countable set. Let $\epsilon >0$. It should be clear that we can choose an increasing sequence $x_1,x_2,...$ such that $x_{k+1}-x_k \in (\frac{\epsilon}{2},\epsilon)$, and $x_k \notin \Omega$ for all $k$. The sequence is, of course, countable, and also closed since all points are isolated. In addition, we must have $\lim_k x_k = \infty$.

Consequently we can create a collection of sequences $k \to x_k^n$ such that $x_{k+1}^n-x_k^n \in (\frac{1}{2^{n+1}},\frac{1}{2^{n}})$ such that $\{x_k^n\}_k \cap \{x_k^m\}_k = \emptyset$ whenever $n \neq m$.

Now let $K_n = \{x_k^n\}_k \cap [n,\infty)$. Then $K_n$ is closed and the sets $\{K_n\}_n$ are pairwise disjoint. By construction, we have $d(K_n,K_m) \leq \frac{1}{2^{\min(n,m)}}$, so presumably is it 'Cauchy'. However, it should also be clear that $K = \emptyset$.

copper.hat
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