Is |x-y| ≤ 2r true for all x,y in a set A containing x₀ with |x-x₀| ≤ r?

[Seda]

Homework Statement

This is the problem stated verbatim. xo is supped to be x with a subscript o.

Suppose that A is a set and there exists xo ε A for which lx-xol ≤ r. Is it necessarily true true that for all x,y εA, we will have lx-yl ≤2r?

Homework Equations

Well, this problem is just supposed to test our ability to formulate proofs, so I can't think of many helpful equations.

I do know lxl < r is the same thing as -r<x<r

The Attempt at a Solution

I can't even think of how to start proving this. In desperation, I tried to think of a counterexample to prove this untrue, but none came to mind. It's also weird that r is such an arbitrary value or variable or something. Nothing is stated in the problem on what it's supposed to represent. I think it means any picked real number but I'm not sure.

Any ideas?

 

[Mathdope]

For clarity, x is supposed to be any element of A correct? If so, then think about using the triangle inequality if it's available.

 

[Seda]

Ok sweet, I completely forgot about the triangle inequality (yes it is available.) However, I am not sure on how to implement it exactly.

So lx+-xol≤ lxl +lxol by the triangle inequality.
lx-yl≤ lxl+lyl ditto.

SO now my equation has become this:

if lxl+lxol≤r
is it true that lxl + lyl≤2r.

Because if I prove that, the desired proof follows?

 

[Mystic998]

No, that's not true. Just because |x - xo| is less than or equal to r doesn't mean that |x| + |xo| is too. Really, think more intuitively about what |x - xo| being less than r says.

 

[Seda]

I'm lost still, mainly beacuse of the confusion between x and xo and what r represents.

Its the difference between a picked value and any other value in a set? sorry If I'm coming across as a total dimwit, but I'm not experiences any revelations here..

Prove that the value of the difference between any two members of a set is always less than twice the value always greater that the difference between any member and one specified member of the same set?

 

[Mathdope]

If |x-x0| < r you can think of it as saying that the disk of radius r about x0 encloses the set. What can you then conclude?

 

[Seda]

R = RADIUS!

I'm an idiot too often, sorry.

So obviously this set can be represented by a circle, where each point is obvious less that a diameter's width away from every other point. diameter = 2r.

I will say that mathdope is a misleading name because you've helped me a ton. Now I just need to ship that explanation up into a mathematical set logic proof rather than a verbal diagram proof, but I thinki should be good to go. Thanks a million

 

[Mathdope]

I will say that mathdope is a misleading name because you've helped me a ton. Now I just need to ship that explanation up into a mathematical set logic proof rather than a verbal diagram proof, but I thinki should be good to go. Thanks a million

A picture should help with the proof. Draw a circle with x and y at random points and x0 at the center. Then try to fit the triangle inequality in with your picture.

 

[Seda]

Sorry to continue to be such a nag, but is there any way I could prove this without referring to the circle at all (my proffessor is crazy about not using diagrams). I understand that lx-yl represents the distance between any two points, which obviously can't exceed 2*r, but then I think I'm assuming the thing I want to prove. Its easy to explain the proof in the context of the circle idea, but without it I seem to stumble even more when I try to just tplainly use inequalities.

Wow, this is the most hopeless I've even felt, and I'm on the dean's list of students too... I really hope I'm just really tired and not normally this stupid.

 

[Mystic998]

Here's a tip that will get you through a good chunk of analysis problems: add 0 inside the absolute value sign in a "creative" way.

 

[Seda]

Hopefully, this will be my last post.

We have x,y as elements of A. Thus we know that

lx-xol≤r

and we know that
ly-xol≤r

based on the question.
So we can add these together to get

lx-xol+ly-xol≤2r

which is the same as

lx-xol + lxo-yl≤2r

By the triangle inequality, we know that

l(x-xo)+(xo-y)l ≤ lx-xol + lxo-yl

So we combine these to get

l(x-xo)+(xo-y)l ≤ lx-xol + lxo-yl ≤ 2r

Thus, l(x-xo)+(xo-y)l ≤ 2r
Finally, lx-yl≤ 2r

Please say that's doable, I want some sleep!
Edit: I have done the bolded step before, even though it's awkward, but my teacher says it's fine in other proofs.

 

[Mathdope]

Looks like you got the jist.

 

[jacobrhcp]

you shouldn't apologize for not knowing something =)