# isometric in metric space

• January 13th 2009, 05:44 PM
aliceinwonderland
isometric in metric space
My textbook says that
For n>=2, consider the set
$A_{n-1} = \{ x = (x_{1},....,x_{n}) \in R^{n} : x_{n}=0\}$ with metric induced by d (usual metric). Then $A_{n-1}$ is a subset of $R_{n}$ and is isometric to $R_{n-1}$ under correspondence

(a) $(x_{1},...,x_{n-1}, 0) \leftrightarrow (x_{1},...,x_{n-1})$

My question is
Isometry should be a bijective function between $A_{n-1}$ and $R_{n-1}$. (a) does not look bijective for me if each $x_{k}$ in $A_{n-1}$ maps to $x_{k}$ in $R_{n-1}$. ( how do we map 0? it should be bijective by definition of isometry)
• January 13th 2009, 08:42 PM
ThePerfectHacker
And isometry does not need to be a bijection.
It just needs to preserve distances. (Wink)
• January 13th 2009, 09:02 PM
aliceinwonderland
According to mathworld, an isometry is defined as follows:

"A bijective map between two metric spaces that preserve distances, i.e,
d(f(x), f(y)) = f(x, y),
where f is the map and d(a,b) is the distance function"

Am I confusing something?
• January 13th 2009, 09:44 PM
ThePerfectHacker
Quote:

Originally Posted by aliceinwonderland
According to mathworld, an isometry is defined as follows:

"A bijective map between two metric spaces that preserve distances, i.e,
d(f(x), f(y)) = f(x, y),
where f is the map and d(a,b) is the distance function"

Am I confusing something?

Not according to this. (Surprised)

This is Mine 11,7:):)th Post!