Metric Space: A Proof of diam(A∪B) ≤ diam(A) + diam(B) | Homework Help

[Maybe_Memorie]

Homework Statement

Consider a metric space (X,d) with subsets A and B of X, where A and B have non-zero intersection. Show that diam(A\bigcupB) \leq diam(A) + diam(B)

Homework Equations

The Attempt at a Solution

A hint would be very much appreciated. Let x\inA, y\inB, z\inA\bigcupB

diam(A\bigcupB) = sup[d(x,y)] for every x,y.
So d(x,y) \leq diam(A\bigcupB)

Either z\inA, or z\inB, or z in both.

Consider firstly z\inA.
d(x,z) \leq diam(A)

If z\inB,
d(z,y) \leq diam(B)

Therefore, d(x,z) + d(z,y) \leq diam(A) + diam(B)

But d(x,y) \leq d(x,z) + d(z,y), hence
d(x,y) \leq diam(A) + diam(B)As far as I've gotten...

 

[hunt_mat]

Examine a point in z\in A\cap B and then examine the distance from x\in A and b\in B and examine the distance from x to y.and apply the definition of diameter.

 

[Maybe_Memorie]

Wait, I have these two results

d(x,y) ≤ diam(A⋃B)

d(x,y) ≤ diam(A) + diam(B)

The latter holds for all x,y, and the maximum value of d(x,y) is diam(A⋃B), which still has to be less than or equal to diam(A) + diam(B).

So, diam(A⋃B) ≤ diam(A) + diam(B)Is my reasoning correct?

 

[hunt_mat]

I think so.

 

[Dick]

Wait, I have these two results

d(x,y) ≤ diam(A⋃B)

d(x,y) ≤ diam(A) + diam(B)

The latter holds for all x,y, and the maximum value of d(x,y) is diam(A⋃B), which still has to be less than or equal to diam(A) + diam(B).

So, diam(A⋃B) ≤ diam(A) + diam(B)


Is my reasoning correct?

That doesn't sound right to me. Where did you get d(x,y)<=diam(A)+diam(B)? Isn't that what you are trying to prove? And where did you use that the intersection of A and B is nonempty? The result is false if that isn't true.

 

[Maybe_Memorie]

That doesn't sound right to me. Where did you get d(x,y)<=diam(A)+diam(B)? Isn't that what you are trying to prove? And where did you use that the intersection of A and B is nonempty? The result is false if that isn't true.

I showed it in my first post, using the triangle inequality. And no, I'm trying to prove that
diam(A\bigcupB) \leq diam(A) + diam(B)

As for using the fact that the intersection is non-empty, if it was empty then the inequality
d(x,y) \leq diam(A) + diam(B) is not necessarily true, but it is non-empty, so it is correct.

 

[Dick]

I showed it in my first post, using the triangle inequality. And no, I'm trying to prove that
diam(A\bigcupB) \leq diam(A) + diam(B)

Ah, indeed you did. Sorry, I didn't read the initial post thoroughly enough.

 

[Maybe_Memorie]

No problem, so is my proof correct?

 

[hunt_mat]

Suppose z\in A\cap B, x\in A and y\in B, the from the triangle inequality:
<br /> d(x,y)\leqslant d(x,z)+d(z,y)<br />
we also know that d(x,z)\leqslant\textrm{diam}(A) and d(z,y)\leqslant\textrm{diam}(B), so we just put these together.

 

[Dick]

No problem, so is my proof correct?

I think you need to be a little more precise. diam(AUB) is the LEAST UPPER BOUND for d(x,y). diam(A)+diam(B) is another upper bound. Just saying they are both upper bounds won't do it. I think that's what you were trying to say with "the maximum value of d(x,y) is diam(A⋃B)" but that's not quite right. There may be no x and y with d(x,y)=diam(AUB).

 

[I like Serena]

I'm afraid you did not apply the triangle property correctly in your opening post.

If you assume that z is in A, then
d(x,z) ≤ diam(A),
but you can not assume that d(z,y) ≤ diam(B).

So you cannot conclude that d(x,z)+d(z,y) ≤ diam(A)+diam(B)


To get your triangle inequality straight you need to use what Dick suggested.

 

[Dick]

I'm afraid you did not apply the triangle property correctly in your opening post.

If you assume that z is in A, then
d(x,z) ≤ diam(A),
but you can not assume that d(z,y) ≤ diam(B).

So you cannot conclude that d(x,z)+d(z,y) ≤ diam(A)+diam(B)To get your triangle inequality straight you need to use what Dick suggested.

Ach, I'm starting mix up who wrote what. But hunt_mat gave the correct argument. I think maybe Maybe_Memory meant to as well but muddled up the cases a bit and didn't mention the triangle inequality.

 

[Maybe_Memorie]

I'm afraid you did not apply the triangle property correctly in your opening post.

If you assume that z is in A, then
d(x,z) ≤ diam(A),
but you can not assume that d(z,y) ≤ diam(B).

So you cannot conclude that d(x,z)+d(z,y) ≤ diam(A)+diam(B)


To get your triangle inequality straight you need to use what Dick suggested.

I'm assuming z is in A, and getting d(x,z) ≤ diam(A)
Then considering the case where z is in B, and getting d(z,y) ≤ diam(B).

Then adding the results.

 

[hunt_mat]

You're considering two different possibilities and you need both possibilities to be true at the same time, you can only do that if z is in the intersection of A and B.

 

[I like Serena]

I'm assuming z is in A, and getting d(x,z) ≤ diam(A)
Then considering the case where z is in B, and getting d(z,y) ≤ diam(B).

Then adding the results.

You cannot combine 2 choices for z in 1 equation.

 

[Maybe_Memorie]

Aaah! Right, I see the problem! So as hunt_mat said

Suppose z\in A\cap B, x\in A and y\in B, the from the triangle inequality:
<br /> d(x,y)\leqslant d(x,z)+d(z,y)<br />
we also know that d(x,z)\leqslant\textrm{diam}(A) and d(z,y)\leqslant\textrm{diam}(B), so we just put these together.

I understand this now.

As for the issue with combining my results to get the desired expression;

d(x,y) \leq diam(A\bigcupB)

d(x,y) \leq diam(A) + diam(B)

diam(A\bigcupB) is the least upper bound for d(x,y) and is contained in d(x,y) so d(x,y) can take the value diam(A\bigcupB)
so diam(A\bigcupB) \leq diam(A) + diam(B)

 

[I like Serena]

You have the right idea, but I'm afraid it's not a proper proof.

First off, you need to split the possibilities for x and y into 4 cases, which can be reduced to 2 cases: x and y in the same set, or x and y in different sets.

Furthermore, there is no guarantee that d(x,y) can take the value diam(A\cupB).
What you can do, is start with your triangle inequality for x,y, and z.
And then take the sup on both sides, yielding the diameters.

 

[Maybe_Memorie]

You have the right idea, but I'm afraid it's not a proper proof.

First off, you need to split the possibilities for x and y into 4 cases, which can be reduced to 2 cases: x and y in the same set, or x and y in different sets.

Furthermore, there is no guarantee that d(x,y) can take the value diam(A\cupB).
What you can do, is start with your triangle inequality for x,y, and z.
And then take the sup on both sides, yielding the diameters.

Okay, if I take d(x,y) \leq d(x,z) + d(z,y), take the sup on both sides,
sup [d(x,y)] \leq sup [d(x,z) + d(z,y)]

Is it correct to say sup [d(x,z) + d(z,y)] \leq sup [d(x,z)] + sup[d(z,y)]
hence sup [d(x,y)] \leq sup [d(x,z)] + sup[d(z,y)] which is what i need?

 

[I like Serena]

Okay, if I take d(x,y) \leq d(x,z) + d(z,y), take the sup on both sides,
sup [d(x,y)] \leq sup [d(x,z) + d(z,y)]

Is it correct to say sup [d(x,z) + d(z,y)] \leq sup [d(x,z)] + sup[d(z,y)]
hence sup [d(x,y)] \leq sup [d(x,z)] + sup[d(z,y)] which is what i need?

It's still slightly more subtle.
What you say is correct btw, but it is not sufficient.

First we have to assume that sup[d(x,y)]=diam(AuB) for x in A and y in B.
(You have to prove the other cases separately.)
Then you can say what you said with an arbitrary z in AnB.
Then you can conclude that sup[d(x,z)] ≤ diam(A) and sup[d(z,y)] ≤ diam(B).
And from this follows your inequality (for this case).