Show that $X$ can be represented as a union of disjoint equivalence classes

Question

Let $X$ be a set. Let $\sim$ be an equivalence relation on $X$. Show that $X$ is the union of disjoint equivalence classes $\{x\}$ for $x \in X$.

What I have tried:

claim: $X = \bigcup_{x\in X} \{ x \}$

indeed, $\{x \} \subset X ;\forall x$ hence $ \bigcup_x \{x \} \subset X$.

Suppose $y \in X$ then here I am stuck - is it OK to write that $y \sim y$ so $y \in \{y \}$? I am not sure I can do that, but if I could it would finish the proof. Any help please

Answer 1

You are in the right way. Since $x\in \{x\}$ (because of reflexivity, as you have said, $x\sim x$) you can conclude that $X \subset \bigcup_{x\in X} \{ x \}.$ As you have shown $\bigcup_x \{x \} \subset X$ you have the equality.

It only remain to show that two different classes are disjoint. Assume $\{x\}\ne \{y\}$ and $\{x\}\cap \{y\}\ne \emptyset.$ Then, there exists $z\in \{x\}\cap \{y\}.$ That, is $x\sim z$ and $z\sim y.$ Becase of transitivity one has $x\sim y.$ So $y\in\{x\}$ from where $\{y\}\subset \{x\}.$ Using a complete analogous argument you have $\{x\}\subset \{y\}.$ Thus, $\{y\}= \{x\}$ which contradicts the assumption that both classes were different.