On 1-1 correspondence and equivalence relation Can someone please explain to me why a 1-1 correspondence between $A$ and $B$ is "clearly" an equivalence relation. 
Thanks for your help.
I got this from the rudin's principle of mathematical analysis. 
 A: Two sets will be in a one-to-one relation if and only if they have the same cardinality.  $$\{(A,B)\mid\exists f~ (f: A\overset{1:1}\mapsto B)\}=\{(A,B)\mid A\equiv B\}$$
It is clear that every set has the same cardinality as itself.   That's reflexivity. $$\forall A~(A\equiv A)$$
Take any two sets.   If the first set has the same cardinality as the second set, then it is clear that second set has the same cardinality as the first.   That's symmetry.$$\forall A\forall B~(A\equiv B\to B\equiv A)$$
Take any three sets.   If the first has the same cardinality as the second, and the second the same cardinality as the third, then it is clear that the first will have the same cardinality as the third.   That's transitivity.$$\forall A\forall B\forall C~((A\equiv B)\wedge (B\equiv A)\to (A\equiv C))$$
So $\{(A,B)\mid\exists f~ (f: A\overset{1:1}\mapsto B)\}$ is clearly an equivalence relation. 
A: The relation is not the same as the bijective function. The definition of the relation is that A~B if there is a bijective function $f:A\to B$. From that definition it is fairly obvious that ~ becomes an equivalence relation on any given set of sets, just using the properties of bijective functions. Namely, that the identity on any set is a bijection, that the inverse of a bijection is a bijection and that the composite of two bijections is again a bijection. 
A: The relation $\sim$ is an equivalence relation because for any sets $A, B$ and $C$: 
$(1)$ There exists a $1-1$ mapping of $A$ onto $A$.
$(2)$ If there exists a $1-1$ mapping of $A$ onto $B$, then there exists a $1-1$ mapping of $B$ onto $A$.
$(3)$ If there exists a $1-1$ mapping of $A$ onto $B$, and there exists a $1-1$ mapping of $B$ onto $C$, then there exists a $1-1$ mapping of $A$ onto $C$.
