# Prove the open ball B(x,r) is a subset of B(x',r') iff d(x,x')<r'-r

1. May 21, 2013

### ghelman

As stated, the problem is to prove B(x,r) is a subset of B(x',r') if and only if d(x,x')<r'-r, where d is a measure function. The proof going one direction seems pretty simple, but I can not figure out the other direction

<=

Assume d(x,x') < r'-r.
Assume a is an element of B(x,r).
Then d(x,x') < r'-r implies d(x,x') + r < r' implies d(x,x') + d(x,a) < r' as d(x,a) < r.
This implies that d(x',a) < r' by use of the triangle inequality, and thus a is an element of B(x',r').

=>

Let a be an element of B(x,r).
Then a is an element of B(x',r').
So d(x,a)<r and d(x',a)<r'.
This implies that d(x',a)-d(x,a) < r'-r

This would almost suggest a use of the reverse triangle inequality, but the inequality is pointing the wrong direction to use here. I am not sure how to proceed.

2. May 21, 2013

### Fredrik

Staff Emeritus
Since this is a textbook-style problem, I'm moving the thread to the calculus & beyond homework forum. You should post questions about textbook-style questions there, and questions about concepts, definitions and theorems in the topology & analysis forum.

Your proof of the first implication is fine, but can be made a bit prettier:

Suppose that $d(x,x')<r'-r$. Let $a\in B(x,r)$ be arbitrary. Then
$$d(a,x')\leq d(a,x)+d(x,x')<r+(r'-r)=r'.$$ To prove the other implication, use that the point at the center of an open ball is an element of the open ball.

3. May 21, 2013

### ghelman

Sorry about putting the thread in the wrong place. I will be sure to put it in the right place next time.

Ok, so since B(x,r) is a subset of B(x',r'), we know that x is an element of B(x',r').
This implies that d(x,x') < r'.

But we need d(x,x') < r' - r

I'm not exactly sure how to proceed from that point. Intuitively, it seems like we want to use the point in B(x,r) that is farthest from x'. However, I have no idea how to find that point.

4. May 21, 2013

### Fredrik

Staff Emeritus
You're right, that was a mistake on my part. I forgot what we were trying to prove, exactly as your post suggests. Unfortunately I don't have time to think this through right now. I'll have another look tomorrow if this isn't solved by then.

5. May 21, 2013

### ghelman

No sweat. Thank you for the help so far!