# Set is Subset of Union/Family of Sets

## 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:

$\ds \forall \beta \in I: S_\beta \subseteq \bigcup_{\alpha \mathop \in I} S_\alpha$

$\blacksquare$