Equivalence Relation on Power Set induced by Intersection with Subset

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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:

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


Then $\alpha$ is an equivalence relation.


Equivalence Class of Empty Set

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

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


Cardinality of Set of Equivalence Classes

Let $A$ be finite with $\card A = n$, where $\card {\, \cdot \,}$ denotes cardinality.

The cardinality of the set of $\alpha$-equivalence classes is given by:

$\card {\set {\eqclass X \alpha: X \in S} } = 2^n$


Proof

Checking in turn each of the criteria for equivalence:


Reflexivity

We have that for all $X \in S$:

$X \cap A = X \cap A$

That is:

$X \mathrel \alpha X$

Thus $\alpha$ is seen to be reflexive.

$\Box$


Symmetry

\(\ds X\) \(\alpha\) \(\ds Y\)
\(\ds \leadsto \ \ \) \(\ds X \cap A\) \(=\) \(\ds Y \cap A\)
\(\ds \leadsto \ \ \) \(\ds Y \cap A\) \(=\) \(\ds X \cap A\)
\(\ds \leadsto \ \ \) \(\ds Y\) \(\alpha\) \(\ds X\)

Thus $\alpha$ is seen to be symmetric.

$\Box$


Transitivity

Let $X \mathrel \alpha Y$ and $Y \mathrel \alpha Z$.

Then by definition:

\(\ds X \cap A\) \(=\) \(\ds Y \cap A\)
\(\ds Y \cap A\) \(=\) \(\ds Z \cap A\)

Hence:

$X \cap A = Z \cap A$

and so by definition of $\alpha$:

$X \mathrel \alpha Z$

Thus $\alpha$ is seen to be transitive.

$\Box$


$\alpha$ has been shown to be reflexive, symmetric and transitive.

Hence by definition it is an equivalence relation.

$\blacksquare$


Sources

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