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

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.

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Fredrik
Staff Emeritus
Gold Member
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.

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.

Fredrik
Staff Emeritus