The Definition of Equivalence Classes of an Equivalence Relation is given as: Suppose $A$ is a set and $R$ is an equivalence relation on $A$. For each element $a$ in $A$, the equivalence class of a, denoted $[a]$ is called the class of a for short, is the set of all elements $x$ in $A$ such that $x$ is related to $a$ by $R$. In symbols, it is $[a] = \{x \in A | xRa \}$. Hence, for all $x \in A$, $x\in[a] \iff xRa$.
Now let $A$ = {0, 1, 2, 3, 4} and define a relation $R$ on $A$ as follows: $R$ = [(0, 0), (0,4), (1, 1), (1, 3), (2, 2), (3, 1), (3, 3), (4, 0), (4, 4)].
Solution: First find the equivalence class of every element of $A$:
[0] = $ \{x \in A | xR0\} = \{0, 4\}$
[1] = $ \{x \in A | xR1\} = \{1, 3\}$
[2] = $ \{x \in A | xR2\} = \{2\}$
[3] = $ \{x \in A | xR3\} = \{1, 3\}$
[4] = $ \{x \in A | xR4\} = \{0, 4\}$
Note that [0] = [4] and [1] = [3]. Thus the distinct equivalence classes of the relation are:
[0, 4], [1, 3], and [2].
My Question is Why isn't the following equivalence classes included as well:
[0] = $ \{x \in A | xR0\} = \{0\}$
[1] = $ \{x \in A | xR1\} = \{1\}$
[3] = $ \{x \in A | xR3\} = \{3\}$
[4] = $ \{x \in A | xR4\} = \{4\}$
Thank you so very much for your help and contributions. I am extremely grateful for your responses. Have a great day.
An equivalence class isn't just a set of things which are all equivalent to each other - it's a set of things which are all equivalent to each other, and which is complete in a sense: anything that you could put into the set, has been.
More precisely, the equivalence class of $x$ is the set of all $y$ which are equivalent to $x$ - so $\{0\}$ isn't an equivalence class because it's missing $4$. Does this make sense?