# Open ball proof

1. Jul 4, 2010

So, I have to prove that in the metric space (R^n, d), where d is the standard Euclidean metric, B(x1, r1) = B(x2, r2) <==> x1 = x2 & r1 = r2.

I finished the proof, but I'm not sure about one step.

Assume B(x1, r1) = B(x2, r2) with x1 = x2. Using the triangle inequality for x1, x and x2, one obtains d(x1, x2) <= d(x1, x) + d(x, x2) < r1 + r2. Now, since this holds for any x1, x and x2, and specially, for any corresponding r1 and r2, i.e. r1 + r2, we conclude that d(x1, x2) = 0, so x1 = x2. Is this correct?

If so, then it is easy to prove the rest - assume B(x1, r1) = B(x2, r2) with x1 = x2 and r1 =/ r2, one easily derives a contradiction.

2. Jul 5, 2010

### Landau

I don't really understand the exercise. Obviously two balls are equal iff they have the same radius and center; that's what "equals" means...
You assume x1=x2 and end up with x1=x2?
You have not explained what x is. Apparently x is both in B(x1,r1) and in B(x2,r2)?

3. Jul 5, 2010

Sorry, it was a mistype. The assumption is x1 =/ x2.

x is any element of B(x1, r1) = B(x2, r2).

4. Jul 5, 2010

### Landau

I see. Then it looks ok, except:

- You don't need to assume x1=/x2: you can just use 'proof by contrapositive' instead of 'proof by contradiction' which I think is always preferable.
-
You have shown that d(x1,x2)< r1+r2. Since r1 and r2 are arbitrary, conclude d(x1,x2)<=0.
In other words, the part "since this holds for any x1, x and x2" is not in place here. (x1 and x2 are not arbitrary, and the fact that x is arbitrary is not relevant).

5. Jul 5, 2010

I understand - I never actually used the fact that x1 =/ x2, so it's not necessary.

OK, thanks a lot!

6. Jul 5, 2010

### Landau

You're welcome!

Yes, WLOG you can assume r2>r1; then any x with
$$d(x_1,x)=d(x_2,x)=\frac{r_2-r_1}{2}$$
would be in B(x2,r2) but not in B(x1,r1).