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A similar question: Distance of two sets and their closest points

The question above, however, defines distance differently. The definition we work under is:

$$\operatorname{dist}(A, B)=\inf\{d(a,b):a \in A, b \in B\}$$

Informally, this must be true. Imagine two open unit discs, $A$ and $B$, centered about $-1$ and $1$ respectively. Notice that the distance $\operatorname{dist}(A, B) = 0$. Even though $A \cap B = \emptyset$, $\partial{A} \cap \partial{B} = \{0, 0\} \neq \emptyset$.

I'm not really sure how to prove this formally, though. Can anyone lend a hint? Is there anything in the informal concept that can be used in proof?

$$\operatorname{dist}(A, B) = 0 \land A \cap B = \emptyset \stackrel{?}{\implies} \partial A \cap \partial B \neq \emptyset$$

Community
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Sean Allred
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  • @DominicMichaelis: The definitions are not equivalent. The one given here always makes sense, while the other one is not well-defined in some metric spaces. The infimum is not always attained, so is not a minimum in general. See the answer below and also the answers to the other question for some examples. – Martin Feb 24 '13 at 16:45
  • @Martin yeah sry you are right. – Dominic Michaelis Feb 24 '13 at 16:48

2 Answers2

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False: Take $A=\mathbb Z^+$, the set of positive integers and $B=\{n+\frac1{2n}: n\in A\}$. Then $A=\partial A, B=\partial B, A\cap B=\emptyset$ and $d(A,B)=0$.

Matemáticos Chibchas
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Another example: Let $A = \left\{(x,y) \in \mathbb{R}^2: xy = 1 \right\}$ and $B = \left\{(x,0): x \in \mathbb{R}\right\}$. $A$ and $B$ are disjoint, closed, and as they have empty interior $\partial A = A$ and $\partial B = B$ while $d((n,0),(n, \frac{1}{n})) = \frac{1}{n}$ so $d(A,B) = 0$.

Henno Brandsma
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  • Hooray for $\mathbb{R}^2$! +1 for easy visual :) – Sean Allred Feb 24 '13 at 17:03
  • @vermiculus Believe or not, the first example that came to my mind was something very similar to Henno's answer: $A={xy=-1,x<0}, B={xy=1,x>0}$, but a "$1$-dimensional" example is more challenging, don't you think? – Matemáticos Chibchas Feb 24 '13 at 19:14
  • @MatemáticosChibchas, agreed on that. However, I'd still say that picturing an example (in $\mathbb{R}^2$) of where it would be true is far easier that picturing an example where it wouldn't, especially when the former has already come to mind.

    I only wish I could accept two answers... (sigh)

     – Sean Allred Feb 24 '13 at 19:57