# Intersection is Commutative/Family of Sets/Proof 1

## Theorem

Let $\family {S_i}_{i \mathop \in I}$ be an indexed family of sets.

Let $\ds I = \bigcap_{i \mathop \in I} S_i$ denote the intersection of $\family {S_i}_{i \mathop \in I}$.

Let $J \subseteq I$ be a subset of $I$.

Then:

$\ds \bigcap_{i \mathop \in I} S_i = \bigcap_{j \mathop \in J} S_j \cap \bigcap_{k \mathop \in \relcomp I J} S_k = \bigcap_{k \mathop \in \relcomp I J} S_k \cap \bigcap_{j \mathop \in J} S_j$

where $\relcomp I J$ denotes the complement of $J$ relative to $I$.

## Proof

We have that both $\ds \bigcap_{j \mathop \in J} S_j$ and $\ds \bigcap_{k \mathop \in \relcomp I J} S_k$ are sets.

Hence by Intersection is Commutative we have:

$\ds \bigcap_{j \mathop \in J} S_j \cap \bigcap_{k \mathop \in \relcomp I J} S_k = \bigcap_{k \mathop \in \relcomp I J} S_k \cap \bigcap_{j \mathop \in J} S_j$

It remains to be demonstrated that $\ds \bigcap_{i \mathop \in I} S_i = \bigcap_{j \mathop \in J} S_j \cap \bigcap_{k \mathop \in \relcomp I J} S_k$.

So:

 $\ds x$ $\in$ $\ds \bigcap_{i \mathop \in I} S_i$ $\ds \leadstoandfrom \ \$ $\ds \forall i \in I: \,$ $\ds x$ $\in$ $\ds S_i$ Definition of Intersection of Family $\ds \leadstoandfrom \ \$ $\ds \forall j \in J: \,$ $\ds x$ $\in$ $\ds S_j$ Definition of Relative Complement $\ds \forall k \in \relcomp I J: \,$ $\, \ds \land \,$ $\ds x$ $\in$ $\ds S_k$ $\ds \leadstoandfrom \ \$ $\ds x$ $\in$ $\ds \bigcap_{j \mathop \in J} S_j$ Definition of Intersection of Family $\, \ds \land \,$ $\ds x$ $\in$ $\ds \bigcap_{k \mathop \in \relcomp I J} S_k$ Definition of Intersection of Family $\ds \leadstoandfrom \ \$ $\ds x$ $\in$ $\ds \bigcap_{j \mathop \in J} S_j \cap \bigcap_{k \mathop \in \relcomp I J} S_k$ Definition of Set Intersection

That is:

$\ds x \in \bigcap_{i \mathop \in I} S_i \iff x \in \bigcap_{j \mathop \in J} S_j \cap \bigcap_{k \mathop \in \relcomp I J} S_k$

The result follows by definition of set equality.

$\blacksquare$