Distance between sets

**Ulysses** · December 21st 2011, 11:51 AM

Hi there. I was working on this problem. And I wanted to know if my demonstration is right, I'm not sure if it's complete.

The problem says: The distance D(A,B) between two nonempty subsets A and B of a metric space (X,d) is defined to be:
$D(A,B)=inf d(a,b),a\in{A},b\in{B}$

Inf denotes the infimus.
Show that D does not define a metric on the power set of X.

Well, so what I considered is that if a belongs to A, and b belongs to B, but a doesn't belongs to B, and b doesn't belongs to A (i.e. A and B has no elements in common), then the distance could never be zero. But I don't know if this is enough to show that it doesn't define a metric.

Anyway, I could show using the sets $A={0,1,2,3...}, B={-1,1,-1,...,(-1)^n},C={0,-1,-2,...}$

That the triangle inequality insn't accomplished, so I have a counter example.

Bye there, and thanks in advance.

**girdav** · December 21st 2011, 12:04 PM

If $A\cap B\neq \emptyset$ , then $D(A,B)=0$ , but we may have this equality even if $A\neq B$ .

The "distance" can be $0$ even if $A$ and $B$ are disjoint: take $X=\mathbb R^2$ , $A=\left\{\left(x,\frac 1x\right),x>0\right\}$ and $B=\mathbb R\times \{0\}$ .

**Ulysses** · December 21st 2011, 12:07 PM

Ok. Then I could use the sets: $A= \left { 0,0,0,0... \right }, B= \left { -1,1,-1,...,(-1)^n \right },C= \left { 0,-1,-2,... \right }$ , with the metric d(x,y)=|x-y| to show that the triangle inequality isn't accomplished, right? I think I got, I've noted it just after I've posted, didn't proved it yet, but I think it can be done easily.

I've defined the metric between the elements of the subsets as d(x,y)=|x-y|, then D(A,B)=1 and D(A,C)=D(C,B)=0, so the triangle inequality isn't accomplished. Is this right?

Thank you.

**Plato** · December 21st 2011, 12:55 PM

Originally Posted by Ulysses

Ok. Then I could use the sets: $A= \left { 0,0,0,0... \right }, B= \left { -1,1,-1,...,(-1)^n \right },C= \left { 0,-1,-2,... \right }$ , with the metric d(x,y)=|x-y| to show that the triangle inequality isn't accomplished, right? I think I got, I've noted it just after I've posted, didn't proved it yet, but I think it can be done easily.

The point is: it is totally unnecessary to do any more.
The problem is done. It fails the "zero test".
So it is not a metric.

**Ulysses** · December 21st 2011, 05:39 PM

With the "zero test" you mean the first proof I approached was right? I thought so, because the null element must always be present in a definite space, right? like in algebra.

**xxp9** · December 21st 2011, 06:29 PM

"zero test" means girdav's first post, saying that, "d(A,B)=0 doesn't mean A=B", this get d fail to be a metric.

**Ulysses** · December 22nd 2011, 03:57 AM

Alright. But is any of my proofs right? I appreciate his work, but is not that intuitive to me. Anyway, I'll take another look at it, but I'd like to know if what I did was right, or if there is any inconsistency on the proofs I've attempted. In the first place, the case I thought of, of A and B having no elements in common, so zero isn't an element of the distance, would that be enough? or doesn't matter at all? and in the other hand, the three sets I've used to proof that it doesn't accomplish the triangle inequality is right?

I see more clearly what he did now. The distance is zero, because x is infinitely close to zero, right?

**FernandoRevilla** · December 22nd 2011, 06:37 AM

Originally Posted by Ulysses

Alright. But is any of my proofs right?

Yes, your counterexample in answer #3 is valid (but please, write correctly the sets!). That is, if $A=\{0\}$ , $B=\{-1,1\}$ and $C=\{0,-1,-2,\ldots\}$ , then $D(A,B)=1$ , $D(A,C)=D(C,B)=0$ hence, $D(A,B)\not\leq D(A,C)+D(C,B)$ , so the triangle inequality is not satisfied.

**Ulysses** · December 22nd 2011, 07:18 AM

Ty Fernando

sorry for that, the thing is that I choose other sets at first, and then "corrected it", because one of the set I choose at first didn't work as I expected, and then I realize I should use the zero set. And the {} didn't know why it wasn't working (I see now I wasn't using the code properly).

Bye there!

**xxp9** · December 22nd 2011, 07:21 AM

use "\{\}" for $\{\}$

**Ulysses** · December 22nd 2011, 04:23 PM

I have a few more doubts,

Originally Posted by girdav

If $A\cap B\neq \emptyset$ , then $D(A,B)=0$ , but we may have this equality even if $A\neq B$ .

The "distance" can be $0$ even if $A$ and $B$ are disjoint: take $X=\mathbb R^2$ , $A=\left\{\left(x,\frac 1x\right),x>0\right\}$ and $B=\mathbb R\times \{0\}$ .

Here, the cartesian product: $B=\mathbb R\times \{0\}$ , which is the product of all the elements of R with zero, is this equal to $B=\{0\}$ I think I understand now, as you defined X in R^2, you must have R^2, right? would it be a mistake to consider an element in R and an element in R^2?

And in the other hand, I wanted to know if the empty set is equal to $B=\{0\}$ , I mean, if the empty set is equal to the set which it's unic element is zero (I think it's not, but one of my companions asked me, and I wasn't completly sure).

Thank you.

**FernandoRevilla** · December 22nd 2011, 11:04 PM

Originally Posted by Ulysses

I have a few more doubts,Here, the cartesian product: ma $B=\mathbb R\times \{0\}$ , which is the product of all the elements of R with zero, is this equal to $B=\{0\}$ I think I understand now, as you defined X in R^2, you must have R^2, right? would it be a mistake to consider an element in R and an element in R^2?

Nothing to do $B=\mathbb R\times \{0\}$ with $C=\{0\}$ . For example, $B$ has infinite elements and $C$ only one, besides, every element of $B$ has the form $(x,0)$ which is different from $0$ .

And in the other hand, I wanted to know if the empty set is equal to $B=\{0\}$ , I mean, if the empty set is equal to the set which it's unic element is zero (I think it's not, but one of my companions asked me, and I wasn't completly sure).

$B=\{0\}$ is a set with only one element: $0$ . The empty set is a set with no elements: $\emptyset=\{\;\}$ . So, $B\neq \emptyset$ .

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