Thread: Diameter of an Open Ball

1. Diameter of an Open Ball

I want to prove that for any ball $\displaystyle B_r(a)$, one has $\displaystyle \text{diam} B_r(a) \leq 2r$

So far I know the following:

$\displaystyle \text{diam} B_r(a) = \sup B_r(a)$. My problem is, what on earth does $\displaystyle \sup B_r(a) = \sup\{\rho(x,a) < r | x,a \in X\}$ actually mean; how do I go about proving that the supremum is indeed less than or equal to 2r?

The statement seems obvious, but the proof for me is a little tricky because of this part. Thanks.

2. Originally Posted by FLT
I want to prove that for any ball $\displaystyle B_r(a)$, one has $\displaystyle \text{diam} B_r(a) \leq 2r$

So far I know the following:

$\displaystyle \text{diam} B_r(a) = \sup B_r(a)$. My problem is, what on earth does $\displaystyle \sup B_r(a) = \sup\{\rho(x,a) < r | x,a \in X\}$ actually mean; how do I go about proving that the supremum is indeed less than or equal to 2r?

The statement seems obvious, but the proof for me is a little tricky because of this part. Thanks.

I think it must be $\displaystyle Diam\, B_r(a)=\sup\{\rho(x,y)\;;\;x,y\in B_r(a)\}$, but then: $\displaystyle \forall\,x,y\in B_r(a)$ :

$\displaystyle \rho(x,y)\le \rho(x,a)+\rho(a,y)=2r\Longrightarrow$ this is also true for the supremum and we're done

Tonio

3. Thanks, I think it was a case of mis-applying the definition then!