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Thread: Hausdroff

  1. #1
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    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} \} $ ?
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    Quote Originally Posted by particlejohn View Post
    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?
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