Union is Associative/Family of Sets
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Theorem
Let $\family {S_i}_{i \mathop \in I}$ and $\family {I_\lambda}_{\lambda \mathop \in \Lambda}$ be indexed families of sets.
Let $\ds I = \bigcup_{\lambda \mathop \in \Lambda} I_\lambda$ denote the union of $\family {I_\lambda}_{\lambda \mathop \in \Lambda}$.
Then:
- $\ds \bigcup_{i \mathop \in I} S_i = \bigcup_{\lambda \mathop \in \Lambda} \paren {\bigcup_{i \mathop \in I_\lambda} S_i}$
Proof
For every $\lambda \in \Lambda$, let $\ds T_\lambda = \bigcup_{i \mathop \in I_\lambda} S_i$.
Then:
\(\ds x\) | \(\in\) | \(\ds \bigcup_{i \mathop \in I} S_i\) | ||||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \exists i \in I: \, \) | \(\ds x\) | \(\in\) | \(\ds S_i\) | Definition of Union of Family | |||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \exists \lambda \in \Lambda: \exists i \in I_\lambda: \, \) | \(\ds x\) | \(\in\) | \(\ds S_i\) | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \exists \lambda \in \Lambda: \, \) | \(\ds x\) | \(\in\) | \(\ds \bigcup_{i \mathop \in I_\lambda} S_i = T_\lambda\) | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds x\) | \(\in\) | \(\ds \bigcup_{\lambda \mathop \in \Lambda} T_\lambda\) |
Thus:
- $\ds \bigcup_{i \mathop \in I} S_i = \bigcup_{\lambda \mathop \in \Lambda} T_\lambda = \bigcup_{\lambda \mathop \in \Lambda} \paren {\bigcup_{i \mathop \in I_\lambda} S_i}$
$\blacksquare$
Also see
Sources
- 1960: Paul R. Halmos: Naive Set Theory ... (previous) ... (next): $\S 9$: Families
- 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 6$. Indexed families; partitions; equivalence relations: Theorem $6.2$