Set is Subset of Union/Family of Sets
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Theorem
Let $\family {S_\alpha}_{\alpha \mathop \in I}$ be a family of sets indexed by $I$.
Then:
- $\ds \forall \beta \in I: S_\beta \subseteq \bigcup_{\alpha \mathop \in I} S_\alpha$
where $\ds \bigcup_{\alpha \mathop \in I} S_\alpha$ is the union of $\family {S_\alpha}$.
Proof 1
Let $x \in S_\beta$ for some $\beta \in I$.
Then:
\(\ds x\) | \(\in\) | \(\ds S_\beta\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(\in\) | \(\ds \set {x: \exists \alpha \in I: x \in S_\alpha}\) | Definition of Indexed Family of Sets | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(\in\) | \(\ds \bigcup_{\alpha \mathop \in I} S_\alpha\) | Definition of Union of Family | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds S_\beta\) | \(\subseteq\) | \(\ds \bigcup_{\alpha \mathop \in I} S_\alpha\) | Definition of Subset |
As $\beta$ was arbitrary, it follows that:
- $\ds \forall \beta \in I: S_\beta \subseteq \bigcup_{\alpha \mathop \in I} S_\alpha$
$\blacksquare$
Proof 2
Let $\beta \in I$ be arbitrary.
Then:
\(\ds \beta\) | \(\in\) | \(\ds I\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \set \beta\) | \(\subseteq\) | \(\ds I\) | Singleton of Element is Subset | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \bigcup \set {S_\beta}\) | \(\subseteq\) | \(\ds \bigcup_{\alpha \mathop \in I} S_\alpha\) | Union of Subset of Family is Subset of Union of Family | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds S_\beta\) | \(\subseteq\) | \(\ds \bigcup_{\alpha \mathop \in I} S_\alpha\) | Definition of Set Union |
So it follows that:
- $\ds \forall \beta \in I: S_\beta \subseteq \bigcup_{\alpha \mathop \in I} S_\alpha$
$\blacksquare$
Sources
- 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $1$: Theory of Sets: $\S 4$: Indexed Families of Sets: Exercise $1 \ \text{(a)}$