Prove the diameter of a union of sets is finite

1. Mar 23, 2014

chipotleaway

1. The problem statement, all variables and given/known data
Prove that the union of a collection of indexed sets has finite diameter if the intersection of the collection is non-empty, and every set in the collection is bounded by a constant A.

3. The attempt at a solution
The picture I have is if they all intersect (and assuming there are infinitely many sets), then all the sets must be 'localized' in some way so that there's some overlap. I'm not yet sure how to make this rigorous though.

2. Mar 23, 2014

Dick

Use the triangle inequality to make a rigorous proof.

3. Mar 23, 2014

chipotleaway

I have the diameter of the union of sets is less than or equal to the sum of the diameters of each set, and that's less than the sum of M (summed over each alpha). How might one go about working in the fact they all intersect?

(If two sets have a maximum diameter of M, and intersect, then I would expect the diameter of the union to be strictly less than the 2M, but if there are infinite sets...)

EDIT: Oh hang on I think I have an idea - x and y in the diameter of the union have to be in either one set or two sets. If in two sets, then since they intersect, then it's less than 2M

4. Mar 23, 2014

Dick

Prove these things you 'expect' to be true using the triangle inequality. And there is at least one point, z, that is contained in every set of the collection.

5. Mar 23, 2014

chipotleaway

Letting d(x,y) be the diameter of the union of the sets, if x,y are in different sets, then d(x,y)≤d(x,z)+d(z,y)≤2M since every set contains z and the diameter of every set is bounded by M. Therefore d(x,y) is finite since 2M is finite.

If x,y are in the same set, then it would be bounded by the diameter of just one set, which is bounded by M.

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I'm a little unsure about using d(x,y) as the diameter of the union of sets and assuming that they're in the sets - should I let the diameter be d(z,w) where z,w are just points in the containing metric space X (which may or may not be in the subsets we're looking)?

6. Mar 23, 2014

Dick

Call your indexed sets $S_i$ where $i \in I$. Then if x and y are elements of the union then $x \in S_i$ and $y \in S_j$ for some $i \in I$ and $j \in I$. Start like that. Now say why you can find a z that is in both sets. Your inequality is fine. Just fill in some more words about what belongs to what.