# Equivalence Relation on Power Set induced by Intersection with Subset

## 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

 $\displaystyle X$ $\alpha$ $\displaystyle Y$ $\displaystyle \implies \ \$ $\displaystyle X \cap A$ $=$ $\displaystyle Y \cap A$ $\displaystyle \implies \ \$ $\displaystyle Y \cap A$ $=$ $\displaystyle X \cap A$ $\displaystyle \implies \ \$ $\displaystyle Y$ $\alpha$ $\displaystyle X$

Thus $\alpha$ is seen to be symmetric.

$\Box$

### Transitivity

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

Then by definition:

 $\displaystyle X \cap A$ $=$ $\displaystyle Y \cap A$ $\displaystyle Y \cap A$ $=$ $\displaystyle Z \cap A$

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.

$\blacksquare$