Equivalence classes – Topics in Algebra Herstein. Property $2$ of an equivalence relation states that if $a \sim b$ then 
$b \sim a$; property $3$ states that if a $a \sim b$ and $b \sim c$ then $a \sim c$. 
What is wrong with the following proof that properties $2$ and $3$ 
imply property $1$ ? Let $a\sim b$; then $b\sim a$, whence, by property $3$ 
(using $a = c$), $a \sim a$. 
 A: I think that for a given $a$ there is not always a $b$ such that $a\sim b$ if you don't have property 1.
A: Consider the following relation on the set $A=\{0,1\}$: the relation is the set of ordered pairs $\{(0,0)\}$.
The relation is clearly symmetric and transitive, but it is not reflexive.
What you're proving is

Suppose $\sim$ is a relation on the set $A$ that is symmetric and transitive; then, if $a\in A$ and there exists $b\in A$ such that $a\sim b$, then $a\sim a$.

The missing property is that the relation is total, that is, for every $a\in A$, there exists $b\in A$ such that $a\sim b$.
Every reflexive relation on $A$ is total. Conversely a total, symmetric and transitive relation on $A$ is reflexive.
A: If $\sim$ is a Binary Relation over a set $X$ and $\exists a,b \in X$ 
such that 
$1.\ a \neq b$
$2. \sim$ is Symmetric and Transitive (i.e. Property $2$ & $3$ hold) 
$3. \ a \sim b$
then indeed $a\sim b \Rightarrow b\sim a\ (symmetry)\Rightarrow a\sim a \ (transitivity)$

However, Reflexivity or (property $1$) states that for all $a \in X$ we must have $a \sim a$. Your approach works for a particular $a$ that too only when we can find a different element $b$ that it is related to. If you can do this for every element of $X$, then Property $2$ and $3$ would imply $1$. In general that is not the case.
