Equivalence Relation on Power Set induced by Intersection with Subset/Equivalence Class of Empty Set

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Theorem

Let $A, T$ be sets such that $A \subseteq T$.

Let $S = \powerset T$ denote the power set of $T$.


Let $\alpha$ denote the relation defined on $S$ by:

$\forall X, Y \in S: X \mathrel \alpha Y \iff X \cap A = Y \cap A$

We have that $\alpha$ is an equivalence relation.


The equivalence class of $\O$ in $S$ with respect to $\alpha$ is given by:

$\eqclass \O \alpha = \powerset {T \setminus A}$


Proof

That $\alpha$ is an equivalence relation is proved in Equivalence Relation on Power Set induced by Intersection with Subset.


We have that:

$\eqclass \O \alpha = \set {X \in S: X \cap A = \O \cap A = \O}$


Thus:

\(\ds X\) \(\in\) \(\ds \eqclass \O \alpha\)
\(\ds \leadsto \ \ \) \(\ds X \cap A\) \(=\) \(\ds \O\) Definition of Equivalence Class
\(\ds \leadsto \ \ \) \(\ds X \cap A\) \(\subseteq\) \(\ds \relcomp T A\) Empty Intersection iff Subset of Complement
\(\ds \leadsto \ \ \) \(\ds X\) \(\subseteq\) \(\ds T \setminus A\) Definition of Relative Complement
\(\ds \leadsto \ \ \) \(\ds X\) \(\in\) \(\ds \powerset {T \setminus A}\) Definition of Power Set
\(\ds \leadsto \ \ \) \(\ds \eqclass \O \alpha\) \(\subseteq\) \(\ds \powerset {T \setminus A}\) Definition of Subset


Then:

\(\ds X\) \(\in\) \(\ds \powerset {T \setminus A}\)
\(\ds \leadsto \ \ \) \(\ds X \cap A\) \(=\) \(\ds \O\) reversing the above argument
\(\ds \leadsto \ \ \) \(\ds X\) \(\alpha\) \(\ds \O\)
\(\ds \leadsto \ \ \) \(\ds X\) \(\in\) \(\ds \eqclass \O \alpha\)
\(\ds \leadsto \ \ \) \(\ds \powerset {T \setminus A}\) \(\subseteq\) \(\ds \eqclass \O \alpha\) Definition of Subset

$\blacksquare$


Sources

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