How do you show that isometry is an equivalence relation among metric spaces? Ok, to start I am new to metric spaces. I have studied equivalence relations in Algebra, but unfamiliar with the e.r. in metric spaces.
Here is my question:
We say that metric spaces (X,$d_X$) and (Y,$d_Y$) are isometric if there is an isometry $f:X \to Y$. Write this as (X,$d_X$) $\backsimeq$ (Y,$d_Y$) and show that $\backsimeq$ is an equivalence relation on the collection of metric spaces.
I know that I need to show the following


*

*$x \backsimeq x$ but don't know what "x" looks like; is it $d_X (a,a)
   \backsimeq d_Y (x,x)$ if so where do "x" come from. Or, how do I
define $\backsimeq$?

*$x \backsimeq y \implies y \backsimeq x$ same problem as above

*$x \backsimeq z$ and $z \backsimeq y \implies x \backsimeq z$.  Which is an even bigger problem since i don't know where "z" comes from.


I am told that if I can show $f^{-1}$ preserves distance, then I am done.
If anyone can tell me what I need to show, I should be good. i.e. what "x" is and how to define $\backsimeq$.
PROOF: (As of right now. I am still over thinking this problem.)
Reflective property can be shown by the identity map. Let $id:X\to X$, then $id$ is bijective and we are done?
Symmetric property want to show $f:X\to Y$ is bijective, which will then show there is inverse to $f$; i.e. $f^{-1}:X\to Y$ which is bijective.
Let $(X,d_X)\backsimeq (Y,d_Y)$, then $f:X\to Y$ is isometry which implies that for every $y\in Y$ $\exists x\in X$ s.t. $f(x)=y$. Since $(X,d_X)$ and $(Y,d_Y)$ are isometric we have $d_X (x_1,x_2)=d_Y (y_1,y_2)=f_Y (f(x_1),f(x_2))$.  Thus, for every $x\in X$ and $y\in Y$ we have $d_X(x_1,x_2)=d_Y(f(x_1),f(x_2))$. Thus, $f$ is bijective, which implies there exists an $f^{-1}$. Therefore, $(Y,d_Y)\backsimeq (X,d_X)$.  
 A: Your objects here are metric spaces. If in algebra you may have met the equivalence relation over $\Bbb Z$ defined by $x\equiv_3 y\iff 3\mid x-y$, and that relation was defined on integers, then this equivalence relation is defined on a collection of metric spaces.
This means that to verify that $x\simeq x$ you actually need to verify that if $(X,d_X)$ is a metric space then there exists an isometry from $X$ to itself (with the same metric!). For example the identity function.
For symmetry you need to show that if there exists $f\colon X\to Y$ which is an isometry between $(X,d_X)$ and $(Y,d_Y)$ then there exists an isometry $g\colon Y\to X$ preserving the same metrics.
Lastly, for transitivity - as before - you have to assume the existence of two isometries between three metric spaces and conclude the existence of a third.

Note that all those are really standard if you have seen proofs of equivalence relations defined by "existence of some structure preserving function".
A: Like you said, "$(X,d_X)\backsimeq(Y,d_Y)$" means that "$(X,d_X)$ is isometric to $(Y,d_Y)$". In particular, "$(X,d_X)\backsimeq(X,d_X)$" means that "$(X,d_X)$ is isometric to $(X,d_X)$". 
In order to prove that the relation $\backsimeq$ is reflexive, you have to show that every metric space $(X,d_X)$ is isometric to itself. In other words, given an arbitrary metric space $(X,d_X)$, you must somehow concoct a function $f:X\to X$ which is an isometry.
To prove that $\backsimeq$ is symmetric means that, given two metric spaces $(X,d_X)$ and $(Y,d_Y)$, and given that $(X,d_X)\backsimeq(Y,d_Y)$, you have to prove that $(Y,d_Y)\backsimeq(X,d_X)$. Now, the fact that $(X,d_X)\backsimeq(Y,d_Y)$ means that there is an isometry $f:X\to Y$. From that you must somehow contrive an isometry $g:Y\to X$
Now for transitivity. You don't have to know where "$z$" comes from, any more than you have to know where "$x$" and "$y$" come from; "$x$", "$y$", and "$z$" are given. You are given three metric spaces $(X,d_X),(Y,d_Y),(Z,d_Z)$, and you are given that $(X,d_X)\backsimeq(Z,d_Z)$ and $(Z,d_Z)\backsimeq(Y,d_Y)$, i.e., you are given isometries $g:X\to Z$ and $h:Z\to Y$; somehow or other you have to manufacture an isometry $f:X\to Y$, thereby showing that $(X,d_X)\backsimeq(Y,d_Y)$.
