1. ## Hausdroff

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

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

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

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

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

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