1. Hausdroff

If $\displaystyle x,y$ are two distinct points in a metric space $\displaystyle X$ then there exists $\displaystyle r > 0$ such that $\displaystyle B_{r}(x)$ and $\displaystyle B_{r}(y)$ are disjoint.

So let $\displaystyle p = d(x,y) > 0$. Choose $\displaystyle r = p/2$. Then $\displaystyle B_{r}(x)$ and $\displaystyle B_{r}(y)$ are disjoint for all $\displaystyle l \leq r$?

Is this right? e.g. $\displaystyle r = \sup A$ where $\displaystyle A = \{l: B_{r}(l) \ \text{and} \ B_{l}(y) \ \text{are disjoint} \}$ ?

2. Originally Posted by particlejohn
If $\displaystyle x,y$ are two distinct points in a metric space $\displaystyle X$ then there exists $\displaystyle r > 0$ such that $\displaystyle B_{r}(x)$ and $\displaystyle B_{r}(y)$ are disjoint.

So let $\displaystyle p = d(x,y) > 0$. Choose $\displaystyle r = p/2$. Then $\displaystyle B_{\color{red}l}(x)$ and $\displaystyle B_{\color{red}l}(y)$ are disjoint for all $\displaystyle l \leq r$?
If they're open balls, yep.

Is this right? e.g. $\displaystyle r = \sup A$ where $\displaystyle A = \{l: B_{r}(l) \ \text{and} \ B_{l}(y) \ \text{are disjoint} \}$ ?
Isn't there a problem with the indices?