Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

A separable, ultrametric space $X$ is given. Does that mean that one immediately gets a countable cover of $X$ consisting of open balls with radius $r>0$ by virtue of separability? (I mean to center a ball with radius $r>0$ at every point of the countable dense subset $U$.)

My question is this: I understand the statement for metric spaces. Does it also hold for ultrametric spaces? Do I have to show anything additionally?

share|improve this question
3  
Ultrametric spaces are, in particular, metric spaces. –  azarel Jan 30 '12 at 23:35
    
The thing about this being an ultrametric space is that every cover by open balls is of this form (every ball of radius $r$ is the ball of radius $r$ about a member of $U$). –  Robert Israel Jan 30 '12 at 23:56
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.