# Metric Space Question

1. Apr 4, 2005

### Ed Quanta

Let e>0 and let Be(r),Be(s) denote the open balls of radius e centered at r,s with respect to the p-adic metric. Prove that if r is an element of Be(s), then Be(s)=Be(r).

Can someone show me how to use the strong triangle inequality to do this?

2. Apr 4, 2005

### PBRMEASAP

Is this what you mean by strong triangle inequality?

d(x,z) <= max{d(x,y), d(y,z)} for all x,y,z in whatever space.

If so, suppose that x is an element of B_e(r). That is, d(x,r) < e. We are given that d(r,s) < e. Then use the triangle inequality to show that d(x,s) < e also. So B_e(r) is contained in B_e(s). Use a similar argument to show B_e(s) is contained in B_e(r).

3. Apr 4, 2005

### Ed Quanta

Yes, that was what I meant by the strong triangle inequality. Why must I use this form of the triangle inequality as opposed to the more general form of the triangle inequality?

And since d(x,s)<=max(d(x,r),d(r,s))

We know d(x,r)<e and d(r,s)<e, so wouldnt this just imply

d(x,s)<2e?

Last edited: Apr 4, 2005
4. Apr 5, 2005

### PBRMEASAP

Because the regular triangle inequality only gets you d(x,s) < 2e, like you showed. This stronger version says that d(x,s) is less than or equal to the larger of these other two "distances" d(x,r) and d(r,s), both of which are strictly less than e. That is, d(x,s) < e. You need this to show that x is also in B_e(s).

5. Apr 5, 2005

### matt grime

And, as a corollary, i you didnt' need the strong one then yo'uve said that all balls of radius e that over lap are equal, and that doesn't seem at all reasonable does it?