Diameter of a union of metric spaces

In summary: Try to follow the same reasoning and you should end up with the same answer.In summary, to prove that a metric space A, which is a union of two subsets B and C of finite diameter, has finite diameter itself, we first establish that the diameter of a metric space M is the supremum of the distances between all points in M. Then, by considering two reference points b and c in B and C respectively, we can find an upper bound for the distance between any two arbitrary points x and y in A. This upper bound does not depend on the specific points chosen, thus proving that A has finite diameter.
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Homework Statement


suppose that a metric space A is a union A = B U C of two subsets of finite diameter. Prove A has finite diameter.


Homework Equations



The Diameter of a metric space M is sup D(a,b) for all a,b in M.

The Attempt at a Solution


Really, no idea where to begin. I just began this subject, and I'm not too well versed in thinking of what these objects really are. If I'm understanding this right, I can imagine having say two disjoint rectangles that are far apart, and these are my B and C, with A being the union of them. Then each B and C may have a diameter, which represents the maximum distance between any two points in them, and part of me wants to take the farthest point of one and connect it to the other, but that seems like that wouldn't always work.

The only thing I can think of is that since both sets have finite diameter, then neither set is infinite. Therefore there must exist a set of points that are the farthest apart...

But like I said, I'm new to this subject (metric spaces) and have no idea where to begin. Any help would be greatly appreciated. (In case it matters, I haven't studied topology, and balls begin in the next section (although I have learned a little about neighborhoods, not sure if that's the same thing)
 
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  • #3
No, I'm sorry. This stuff is completely foreign to me. I have no idea how I can show this.
 
  • #4
I suggest that you start by picking two reference points, say b and c, where [itex]b \in B[/itex] and [itex]c \in C[/itex]. Being points in a metric space, there is by definition some finite distance, say d(b,c), between them.

Now take arbitrary points x and y in [itex]B \cup C[/itex]. See if you can find an upper bound for d(x,y) that doesn't depend on x and y.
 
  • #5
You've got the solution in the thread I sent you! It's shwon in problem 3.
 

1. What is the definition of a union of metric spaces?

A union of metric spaces is the set that contains all the elements from each of the individual metric spaces. It is denoted by the symbol ∪ and is used to combine the elements of two or more metric spaces into one set.

2. How is the diameter of a union of metric spaces calculated?

The diameter of a union of metric spaces is the maximum distance between any two points in the combined set. It is calculated by taking the supremum (or the least upper bound) of all the distances between points in the union of metric spaces.

3. Can the diameter of a union of metric spaces be infinite?

Yes, the diameter of a union of metric spaces can be infinite if one or more of the individual metric spaces have an infinite diameter. In this case, the union of metric spaces will also have an infinite diameter.

4. What is the relationship between the diameter of a union of metric spaces and the individual metric spaces?

The diameter of a union of metric spaces is always greater than or equal to the diameter of each individual metric space. This is because the union of metric spaces contains all the elements from each individual metric space, so the maximum distance between any two points in the union will be at least as large as the maximum distance in each individual metric space.

5. Can the diameter of a union of metric spaces be smaller than the individual metric spaces?

No, the diameter of a union of metric spaces can never be smaller than any of the individual metric spaces. This is because the union contains all the elements from each individual metric space, so the maximum distance between any two points in the union will always be greater than or equal to the maximum distance in each individual metric space.

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