Element in its own Equivalence Class

From ProofWiki

Jump to: navigation, search

Theorem

Let $\mathcal R$ be an equivalence relation on a set $S$.

Then every element of $S$ is in its own $\mathcal R$-class:

$\forall x \in S: x \in \left[\!\left[{x}\right]\!\right]_\mathcal R$

	$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle \forall x \in S: \left({x, x}\right)$	$\in$	$\displaystyle \mathcal R$	$\displaystyle $	$\displaystyle $	$\displaystyle $	Equivalence relation is reflexive
	$\displaystyle $	$\displaystyle \implies$	$\displaystyle $	$\displaystyle x$	$\in$	$\displaystyle \left[\!\left[{x}\right]\!\right]_\mathcal R$	$\displaystyle $	$\displaystyle $	$\displaystyle $	Definition of Equivalence Class

$\blacksquare$

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \forall x \in S: \left({x, x}\right)\)	\(\in\)	\(\displaystyle \mathcal R\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Equivalence relation is reflexive
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle x\)	\(\in\)	\(\displaystyle \left[\!\left[{x}\right]\!\right]_\mathcal R\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Definition of Equivalence Class