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let X,d be a metric space d(x,A) = inf{d(x,a):a of A} d(A,B) = inf{d(a,b):a of A, b of B}

(i) prove that d(x,A) = 0 iff x is an element of A bar for this I came to the conclusion that the above statement isn't true, assuming A bar is the compliment of A then if x is and element of A then d(x,x) = 0 is an element of {d(a,b):a of A, b of B} and it's infimum and is therefore d(x,A)=0 for x an element of A not A bar. was it a trick question or am I wrong?

(ii) Give an example of closed, disjoint subsets A and B of the plane R^2 for which d(A,B) = 0. does such an example exist? since A and B are closed they contain there boundary points, so d(A,B) = the distance between their closest boundary points since they are disjoint those aren't the same point and therefore it isn't equal to 0.

was this a trick question, or is my logic off?

Damien King-Acevedo
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    I think $\bar{A}$ means the closure of $A$, not the complement. – Dan Shved Oct 29 '13 at 17:20
  • In part (ii), your mistake is in assuming that there are "closest boundary points". Might be there are arbitrarily close pairs of points, one from $A$ and one from $B$, but no single closest pair. – Dan Shved Oct 29 '13 at 17:21
  • Hint: You cannot get an example of (ii) where $A$ our $B$ are bounded (because they would then be compact). So you need to look elsewhere. – Marc van Leeuwen Oct 29 '13 at 17:22
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    Not a trick question. Look at curve $xy=1$ and the $x$-axis. Or let $A$ be the set of positive integers, and $B$ the set of numbers of the form $n+\frac{1}{2^n}$ where $n$ ranges over the positive integers. (In principle these are subsets of $\mathbb{R}^2$, but we can identify them with certain points on the $x$-axis). – André Nicolas Oct 29 '13 at 17:23
  • thanks a lot, A bar being the closure makes a whole lot more sense. On the second part I was trying to define the sets as areas of R^2, your examples were very helpful. Thanks. – Damien King-Acevedo Oct 29 '13 at 17:33

2 Answers2

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Think perhaps of the sets $A=\mathbb{N}$ and $B=\mathbb{N}+ \pi/n= \{n+\frac{\pi}{n} | n \in \mathbb{N}\}$. Then there always are points of A and B which are arbitrarily close, but none of them is a limit point of the other set.

Vladhagen
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What if points between A and B get arbitrarily close but never touch? So in one spot the distance is 1. In another spot far away the distance is 1/2. Etc.

Jonathan Aronson
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