# Thread: Metric spaces and Sets...

1. ## Metric spaces and Sets...

Hello!
Here's a problem I have:
Let the distance from set A to set B be defined as:
dist(A,B) = inf {d(x,y) : x is in A, y is in B} (d(x,y) is a metric)
A & B are closed sets in R^n. A is bounded.
Prove: dist(A,B)>0 <=> A and B are disjoint sets.
Then they ask: Is the sentence still correct if A isn't bounded?

Now, I thought I had a pretty neat proof, but I didn't use the fact that A is bounded. I didn't even use the fact that A and B are closed sets!
I shall shortly detail my proof, and you guys can tell me if I'm missing something:
assume dist(A,B)>0.
let's assume (by contraction) that A and B are not disjoint, i.e., there's a z so that z is in A and in B.
So, d(z,z) is in {d(x,y) : x in A, y in B). but d(z,z) = 0 , so (needs formal proof, but), inf{d(x,y): x in A, y in B) = 0 => dist (A,B) = 0. Contraction.
The other way is pretty similar.

What am I missing? Why do I need A and B to be closed sets? Why do I need A to be bounded?

Thanks!!!!
Tomer.

2. Oh, I realized now that the other direction of the sentence probably needs this extra data. I forgot that if I (contradically) assume that dist(A,B) = 0, it doesn't mean that 0 is in {d(x,y) : x in A, y in B).
I need to work on that...

Thanks anyway!

3. I doubt that boundeness has any thing to do with it.
Consider A=(0,1) & B=[1,2]. The D(A;B)=0 but $\displaystyle A \cap B = \emptyset$.
Note that closure changes the argument.

4. If A and B are both closed, it's still possible to have dist(A,B) = 0, but only if A and B are unbounded. For example, in R^2 (usual metric), A could be the x-axis, and B the set {(x,1/x): x>0}.

5. Thank you for the replies!

6. ## Still don't get it...

Hi. I'd really appreciate it if someone would check out this next proof for the problem discussed here. It seems to be correct, but I didn't use the fact that A and B are closed, nor that A is bounded.... which must be wrong.
What the problem here?

One direction - I've proved above.
Assume A and B are disjoint.
We need to prove that dist(A,B)>0
Let's assume, by contraction, that dist(A,B) = 0. => to every e>0, exists x in A and y in B so that d(x,y)<e. But that means that x=y. So, x is in A, and x is in B, which means d(x,x) is in {d(x,y) | x in A, y in B}. but d(x,x) = 0. That's a contradiction to the fact that dist(A,B)>0.

What am I missing? :-\

7. Originally Posted by aurora
We need to prove that dist(A,B)>0
Let's assume, by contraction, that dist(A,B) = 0. => to every e>0, exists x in A and y in B so that d(x,y)<e. But that means that x=y.
That is not true. That does not follow at all.
If A is closed & bounded is it compact?
Can you get a sequence of points in A that converges to a limit point of B?
If that happens can A & B be disjoint?

8. Thank you.
I'm trying to work it out but it seems I'm stuck.

Ok, I wrote a huge speech regarding why my proof is incorrect, and then I realized that d(x,y)<e for specific x,y's... stupid me

However, I still seem to be unaware of the correct proof.
Sure, A is compact. I doesn't tell me much though - I've hardly learned anything about compact groups. I think the only sentence that might be useful here (though I don't know how), is that in compact groups (in R^n), each series of points in A, has a sub-series that converges to a point in A.
If I could find a sequence of points in A that converges to a limit point of B, that would mean that this point is an accumulation point of A, and because A is closed, that would mean that this point is both in A and in B, and that's of course a contradiction.
But where do I start from? What's my starting anchor? How's that got to do with dist(A,B)?
I'm sorry I don't see the answer, this field is very new to me and the homework needs to be done :-\
Just a little push, or a relavent (basic) sentence, must help...

9. Frankly, I have not tried to reply for trying to give you a method that fits with what you have said is your understanding of these ideas.
Given that B is a closed set if x is not in B then D(B;x)>0.
So let’s suppose that A & B are disjoint closed sets and D(B;A)=0.
$\displaystyle \varepsilon _1 = 1,\quad \left( {\exists a_1 \in A} \right)\left[ {0 < D\left( {B;a_1 } \right) < \varepsilon _1 } \right]$.
Let $\displaystyle \varepsilon _2 = \min \left\{ {\frac{{D\left( {B;a_1 } \right)}}{2},\frac{1}{2}} \right\}$ then $\displaystyle \left( {\exists a_2 \in A} \right)\left[ {D(B;a_2 ) < \varepsilon _2 } \right]$.
Do you see that $\displaystyle a_1 \ne a_2$??
Then can you continue this process?
Because A is compact, then this sequence has a limit point.
Would that point be in the closure of B?