Intersection is Commutative/Family of Sets/Proof 1
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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.
$\blacksquare$
Sources
- 1960: Paul R. Halmos: Naive Set Theory ... (previous) ... (next): $\S 9$: Families