# Open balls in metric spaces

• Feb 12th 2009, 11:06 AM
pkr
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.
• Feb 12th 2009, 11:19 AM
Plato
Quote:

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 $\displaystyle p \notin B_1 (0)$?
• Feb 12th 2009, 11:21 AM
Moo
Hello,
Quote:

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 :
$\displaystyle \frac{|x|}{1+|x|}<1$
$\displaystyle |x|<1+|x|$
which gives $\displaystyle 1>0$, for any x. It means that the inequality is true for any x.
So the open ball is the whole set.

Quote:

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 $\displaystyle B_r(x)=\{y \in X ~:~ d_1(x,y)<r\}$ or $\displaystyle =\{y \in X ~:~ d(x,y)<r\}$ ?

Is d the same as above ?
• Feb 12th 2009, 11:32 AM
pkr
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.
• Feb 12th 2009, 12:46 PM
pkr
I would assume that's true, but i'm confused about the whole "Let x ≠ 0"
• Feb 12th 2009, 01:23 PM
Plato
Here is a fix, I hope.
Use the notation $\displaystyle B_r (x)$ for a ball in the original space with metric $\displaystyle d$, and $\displaystyle \mathbb{B}_r (x)$ as a ball is the new space with metric $\displaystyle d_1$.
For $\displaystyle 0<r\le d(x,o)$ then $\displaystyle \mathbb{B}_r (x) = \{x\}$.
For $\displaystyle r> d(x,o)$ then $\displaystyle \mathbb{B}_r (x) = B_{r - d(x,o)}(o)$.