# Open balls in metric spaces

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• February 12th 2009, 12:06 PM
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.
• February 12th 2009, 12:19 PM
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 $p \notin B_1 (0)$?
• February 12th 2009, 12:21 PM
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 :
$\frac{|x|}{1+|x|}<1$
$|x|<1+|x|$
which gives $1>0$, 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 $B_r(x)=\{y \in X ~:~ d_1(x,y) or $=\{y \in X ~:~ d(x,y) ?

Is d the same as above ?
• February 12th 2009, 12:32 PM
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.
• February 12th 2009, 01:46 PM
pkr
I would assume that's true, but i'm confused about the whole "Let x ≠ 0"
• February 12th 2009, 02:23 PM
Plato
Here is a fix, I hope.
Use the notation $B_r (x)$ for a ball in the original space with metric $d$, and $
\mathbb{B}_r (x)$
as a ball is the new space with metric $d_1$.
For $0 then $\mathbb{B}_r (x) = \{x\}$.
For $r> d(x,o)$ then $\mathbb{B}_r (x) = B_{r - d(x,o)}(o)$.