Stuck on Rudin's Real Analysis 2.3 Definition regarding Cardinality and Equivalence Relations I have been self-studying Real Analysis and drudging through Rudin's Principles of Mathematical Analysis. I have hit a bit of a stumbling block at definition 2.3, where he seems to state that sets $A$ and $B$ are equivalent (that there is an equivalence relation on them) if the sets have the same cardinal number.
I am failing to see how equivalence follows from cardinal equivalence, but I know I must be mistaken or misreading/misunderstanding this (as it's obvious Rudin cannot be wrong, so therefore, the error lies within my feeble mind).
Here is the definition:

2.3 Definition
  If there exists a 1-1 mapping of $A$ onto $B$, we say that $A$ and $B$ can be put in a 1-1 correspondence, or that $A$ and $B$ have the same cardinal number, or briefly, that $A$ and $B$ are equivalent, and we write $A \sim B$. This relation clearly has the following properties:
  
  
*
  
*It is reflexive: $A \sim A$.
  
*It is symmetric: If $A \sim B$,  then $B \sim A$. 
  
*It is transitive: If $A \sim B$ and $B \sim C$, then $A \sim C$.
  

My understanding (likely misunderstanding) is as follows:
For nonempty sets $A$ and $B$, let there be a bijection from $A$ to $B$. It follows that $|A| = |B|$. For example, let $A = \{0, 2, 4\}$ and $B = \{1, 3, 5\}$. Clearly $|A| = |B| = 3$. Let $f: A \rightarrow B$ be defined by $f(n) = n + 1$. Clearly, $f$ is bijective. However, it does not seem to follow from $|A| = |B|$ that $A \sim B$. 
For a relation $R$ from $A$ to $B$ to be equivalent, it must satisfy reflexivity, symmetry, and transitivity. In my proofs text I can only find examples of these relations on a single set $A$ rather than between two sets. I am convinced herein lies my confusion, because given my example of two sets above of equal cardinality, neither reflexivity, symmetry, nor transitivity hold.
Any pointers in the right direction would be greatly appreciated at this moment of brain meltdown.
Thanks in advance!!!
 A: I think you're misunderstanding the concept of the equivalence relation.
An equivalence relation is not like other mathematical concepts with which you are familiar. To really understand what an equivalence relation is, it's important that you first understand the general notion of a relation.
We can say that a relation is a rule for specifying relationships between mathematical objects. Pretty much any statement that you can make regarding the way in which such objects are similar or dissimilar can be classified as a relation. For example, for two sets $A$, $B$, the statement $A$ is a subset of $B$ is a relation. It describes a relationship between the two sets, and we can say that $A$ and $B$ are related in terms of this relation. Now, this relation is both reflexive and transitive because for any set we have $A \subseteq A$ and if $A \subseteq B$ and $B \subseteq C$ then $A \subseteq C$. However, the statement $A$ is a subset of $B$ does not define an equivalence relationship between sets because it does not satisfy the requirement of reflexivity.
Now let's take a look at the example you gave. In your example, you are defining that $A$ is related to $B$ if $A$ and $B$ have the same cardinality. This is indeed an equivalence relation because:


*

*$|A| = |A|$ for any set $A$.

*If $|A| = |B|$ then $|B| = |A|$ for any sets $A$, $B$.

*If $|A| = |B|$ and $|B| = |C|$ then $|A| = |C|$ for any sets $A$, $B$, $C$.


In terms of bijections, $|A| = |B|$ if there exists a bijection between the elements of $A$ and $B$. You gave the function $f$ as an example of a bijection showing that $A \sim B$. To show that $B \sim A$ you would use the bijection $f^{-1}: B\to A$. Now say we have three sets, $A = \{0, 2, 4\}$, $B = \{1, 3, 5\}$, and $C = \{3, 5, 7\}$. We know that $A \sim B$ using $f(n) = n + 1$ and we know $B \sim C$ using $g(n) = n + 2$ so $A \sim C$ using $g(f(n)) = n + 3$.
Hope this helps!
