Is an equivalence class of an equivalence relation , an equivalence relation itself? So , I have been revising my lecture notes on relations . When I read the definition of an equivalence class of an element a of a relation R, I came forward those bullets:

*

*reflective: $ a \in [a]_R$

*symmetric: if $b \in [a]_R $ then $a \in [b]_R$

*transitive: if $ b \in [a]_R$ and $c \in [a]_R$ then $(b,c) \in R$ ( I don't quite get that : If b is related to a and c is related to a , why it is sure that b is related to c?)


So this is supposed to show us that an equivalence class is an equivalence relationship itself?
As far as I have undestood theory , a relation basically defines a set ( of elements that satisfy whatever this relation defines). An equivalence class, also defines a set

 A: Given a set $A$ and a relation $R$ on $A$, $R$ does not define a set; rather $R$ is a set. More precisely, it is a subset of $A\times A$.
If it turns out that $R$ is an equivalence relation, then, for each $a\in A$, you define the equivalence classe of $a$ as:$$[a]=\{b\in A\mid a\mathrel Rb\}.$$Note that $[a]$ is a subset of $A$, whereas $R$ is a subset of $A\times A$. And the set of all equivalence classes is a subset of $\mathcal P(A)$.
A: Intuitively you can think of an equivalence class as a set of things you can transform between. So if $a \in [b]_{R}$ then you can transform from $a$ to $b$ (and vice versa due to the symmetric property) which I will denote by $a\rightarrow b$. Since you can perform the transform $a\rightarrow b$ and $c \rightarrow a$ (again symmetry property) then you can combine them to get:
$$ c\rightarrow a \rightarrow b$$
and therefore you can transform from $c$ to $b$ (and vice versa) and so $b$ is in the equivalence class of $c$. This means there is a relation between $b$ and $c$ and so $(b,c)\in R$.
