Equivalent Characterizations of Finer Equivalence Relation

Theorem
Let $X$ be a set.

Let $\equiv$ and $\sim$ be equivalence relations on $X$.


 * 1) $\equiv$ is finer than $\sim$:
 * $\forall x, y \in X : x \equiv y \implies x \sim y$
 * 1) The graph of $\equiv$ is contained in the graph of $\sim$.
 * 2) Every $\equiv$-equivalence class is contained in a $\sim$-equivalence class.
 * 3) Every $\sim$-equivalence class is saturated under $\equiv$.

1 implies 2
By definition of subset, $\map \TT \equiv \subseteq \map \TT \sim$.

1 implies 3
Let $x \in X$.

Then:

By definition of subset, $\eqclass x \equiv \subseteq \eqclass x \sim$.

3 implies 1
For this proof and the next, we use this result implied by $3$:

Let $y \in X$.

$3$ implies:
 * $\exists z \in X: \eqclass y \equiv \subseteq \eqclass z \sim$

By Element in its own Equivalence Class:
 * $y \in \eqclass y \equiv \subseteq \eqclass z \sim$
 * $y \in \eqclass y \sim$

By contrapositive of Equivalence Classes are Disjoint:
 * $\eqclass y \sim = \eqclass z \sim$

Hence $\eqclass y \equiv \subseteq \eqclass y \sim$.

Thus:

3 implies 4
Let $x \in X$.

so $\eqclass x \sim$ is a union of equivalence classes under $\equiv$.

Hence $\eqclass x \sim$ is saturated under $\equiv$.

4 implies 3
Let $x \in X$.

Then $\eqclass x \sim$ is saturated under $\equiv$:
 * $\displaystyle \exists S \subseteq X: \bigcup_{y \mathop \in S} \eqclass y \equiv = \eqclass x \sim$

By Element in its own Equivalence Class:
 * $x \in \eqclass x \sim$
 * $x \in \eqclass x \equiv$

Hence by definition of Union of Family:
 * $\exists z \in S: x \in \eqclass z \equiv$

By contrapositive of Equivalence Classes are Disjoint: