Thread: Open balls in metric spaces

d(x,y) = |x-y| / 1+|x-y|
Describe B1(0), open ball of centre 0 and radius 1;

so i've got (|x| / 1+|x|) < 1
Is that enough of a description? Not sure what I could say about Br(a)

And (X,d) a metric space, fix a point o in X
let d1(x,y)= d(x,o) + d(o,y)
if x ≠ y and d1(x,x)=0

Let x ≠ 0, describe Br(x) w.r.t d1 for all possible r

Not sure at all with that one, any hints would be appreciated.

Originally Posted by pkr

d(x,y) = |x-y| / 1+|x-y|
Describe B1(0), open ball of centre 0 and radius 1

Here is a question for you.
Does there exist a point p in this metric space such that ?

Hello,

Originally Posted by pkr

d(x,y) = |x-y| / 1+|x-y|
Describe B1(0), open ball of centre 0 and radius 1;

so i've got (|x| / 1+|x|) < 1
Is that enough of a description? Not sure what I could say about Br(a)

Maybe you can try to simplify !
1+|x|>0 for any x. So you can multiply both sides by 1+|x|, without changing the inequality :


which gives , for any x. It means that the inequality is true for any x.
So the open ball is the whole set.

And (X,d) a metric space, fix a point o in X
let d1(x,y)= d(x,o) + d(o,y)
if x ≠ y and d1(x,x)=0

Let x ≠ 0, describe Br(x) w.r.t d1 for all possible r

Not sure at all with that one, any hints would be appreciated.

So is or ?

Is d the same as above ?

For the 1st question I assumed it to be the whole metric space but wasn't sure if it was possible, couldn't find any notes on that so many thanks.

The second part is a different question, where (X,d) is simply any metric space.

I would assume that's true, but i'm confused about the whole "Let x ≠ 0"

Here is a fix, I hope.
Use the notation for a ball in the original space with metric , and as a ball is the new space with metric .
For then .
For then .