Proof:⟦a⟧=⟦b⟧ if and only if aRb

From CS2800 wiki
If [math]R [/math] is an equivalence relation on [math]A [/math], then [math]\href{/cs2800/wiki/index.php/%E2%9F%A6a%E2%9F%A7}{⟦a⟧}_R = ⟦b⟧_R [/math] if and only if [math]\href{/cs2800/wiki/index.php/XRy}{aRb} [/math].
Proof:
Suppose first that [math]\href{/cs2800/wiki/index.php/%E2%9F%A6a%E2%9F%A7}{⟦a⟧}=\href{/cs2800/wiki/index.php/%E2%9F%A6a%E2%9F%A7}{⟦b⟧} [/math]; we will show that [math]\href{/cs2800/wiki/index.php/XRy}{aRb} [/math]. Since [math]R [/math] is reflexive, we have [math]bRb [/math]. Thus [math]b \href{/cs2800/wiki/index.php/%5Cin}{\in} \href{/cs2800/wiki/index.php/%E2%9F%A6a%E2%9F%A7}{⟦b⟧} [/math]. Since [math]\href{/cs2800/wiki/index.php/%E2%9F%A6a%E2%9F%A7}{⟦a⟧}=\href{/cs2800/wiki/index.php/%E2%9F%A6a%E2%9F%A7}{⟦b⟧} [/math], this means [math]b \href{/cs2800/wiki/index.php/%5Cin}{\in} \href{/cs2800/wiki/index.php/%E2%9F%A6a%E2%9F%A7}{⟦a⟧} [/math], which by definition means [math]\href{/cs2800/wiki/index.php/XRy}{aRb} [/math], as required.


Now, suppose aRb; we will show that [math]\href{/cs2800/wiki/index.php/%E2%9F%A6a%E2%9F%A7}{⟦a⟧}=\href{/cs2800/wiki/index.php/%E2%9F%A6a%E2%9F%A7}{⟦b⟧} [/math]. Since [math]aRb [/math], we have [math]b \href{/cs2800/wiki/index.php/%5Cin}{\in} \href{/cs2800/wiki/index.php/%E2%9F%A6a%E2%9F%A7}{⟦a⟧} [/math]. Since [math]R [/math] is reflexive, we have [math]bRb [/math] so [math]b \href{/cs2800/wiki/index.php/%5Cin}{\in} \href{/cs2800/wiki/index.php/%E2%9F%A6a%E2%9F%A7}{⟦b⟧} [/math]. Since equivalence classes form a partition, and [math]\href{/cs2800/wiki/index.php/%E2%9F%A6a%E2%9F%A7}{⟦a⟧} [/math] and [math]\href{/cs2800/wiki/index.php/%E2%9F%A6a%E2%9F%A7}{⟦b⟧} [/math] overlap, we conclude that [math]\href{/cs2800/wiki/index.php/%E2%9F%A6a%E2%9F%A7}{⟦a⟧}=\href{/cs2800/wiki/index.php/%E2%9F%A6a%E2%9F%A7}{⟦b⟧} [/math], as required.


Note that you can prove this result without resorting to the fact that equivalence classes form a partition, but you would just be repeating a lot of the work done in that proof.