Proving Disjoint Closed Sets in Metric Space: A Compact Set

In summary: This guarantees that the sequences converge to the same point, but it doesn't say anything about the distance between those points.
  • #1
giantsfan000
3
0

Homework Statement



Suppose A and B are disjoint closed sets in the metric space X and assume
in addition that A is compact. Prove there exists ∆ > 0 such that for all
a ∈ A, b ∈ B, d(a, b) ≥ ∆

2. The attempt at a solution

I really don't have an attempt at a solution because I am 100% completely stuck. I just don't know where to start. I'm not looking for any answers or anything, if someone could just tell me the first step or so, or point me in the right direction, I'd be much obliged.

Thanks!
 
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  • #2
A proof by contradiction is a good place to look. What if d(a,b) can be arbitrarily small?
 
  • #3
I appreciated your prompt response to my question about topology. Thank you very much for that! Proof by contradiction does make sense. I hope you don't mind me asking a follow up question. Would the contradiction be that there is not an delta > 0 st d(a,b) > delta or there exists a delta > 0 st d(a,b) < delta. I really appreciate the help. This problem was assigned as a review, and I keep trying to do it, but it's really kicking my butt and I just keep getting stuck over and over again! I guess I just need to figure out my proper proof by contradiction statement and then see if I can somehow figure it out from there.
 
  • #4
If there exists no delta as described in the problem, then for any value delta there exists a and b such that d(a,b)<delta
 
  • #5
Thank you so much for your help office shredder. I really do hate to ask a follow up again because I know you don't want to do too much of the problem for me, but I just feel so stupid when I try to do this problem. So I've got my statement set up that I want to contradict, but how exactly do I go about contradicting this? I know I need to make use of the fact that one of the sets is compact because if they were only both closed then I don't think this statement holds. So I want to use that they're disjoint, ones compact, ones closed, but how exactly would I do this? And again, I'm sorry to be asking so much, I usually try to just think about it until I figure stuff out, but this problem has just been kicking my butt all day long :(
 
  • #6
Use the fact that you can find points in A and B that are arbitrarily close to construct a sequence of points in A and a sequence of points in B that do something special.

A common thing to do in situations like this is to consider finding points a and b for each choice of n with delta=1/n
 

Related to Proving Disjoint Closed Sets in Metric Space: A Compact Set

1. What is a metric space?

A metric space is a mathematical construct that consists of a set of elements and a function that defines the distance between those elements. The function is known as a metric and must satisfy certain properties, such as being non-negative and symmetric.

2. What does it mean for two sets to be disjoint?

Two sets are disjoint if they have no elements in common. In other words, there is no overlap between the two sets and they are completely separate from each other.

3. What is a compact set?

A compact set is a subset of a metric space that is closed and bounded. This means that the set contains all of its limit points and is contained within a finite distance from any given point in the metric space.

4. How do you prove that two sets are disjoint and closed in a metric space?

To prove that two sets are disjoint and closed, you must first show that they have no elements in common. Then, you must show that their complements (the elements that are not in the sets) are open in the metric space. This can be done by showing that for any point in the complement, there exists a small open ball around that point that is contained entirely within the complement.

5. Why is proving the disjointness and closedness of sets important in a metric space?

Proving the disjointness and closedness of sets is important in a metric space because it allows us to make conclusions about the behavior of the sets within the space. For example, if two sets are disjoint, we know that they do not interact with each other and can be considered separately. Additionally, if a set is closed, we know that it contains all of its limit points and is well-behaved within the metric space.

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