Set is Subset of Union

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Theorem

The union of two sets is a superset of each:

$S \subseteq S \cup T$
$T \subseteq S \cup T$


General Result

Let $S$ be a set.

Let $\mathcal P \left({S}\right)$ be the power set of $S$.

Let $\mathbb S \subseteq \mathcal P \left({S}\right)$.


Then:

$\forall T \in \mathbb S: T \subseteq \bigcup \mathbb S$


Set of Sets

Let $\mathbb S$ be a set of sets.


Then:

$\displaystyle \forall T \in \mathbb S: T \subseteq \bigcup \mathbb S$


Indexed Family of Sets

In the context of a family of sets, the result can be presented as follows:

Let $\left \langle{S_\alpha}\right \rangle_{\alpha \mathop \in I}$ be a family of sets indexed by $I$.


Then:

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

where $\displaystyle \bigcup_{\alpha \mathop \in I} S_\alpha$ is the union of $\left \langle{S_\alpha}\right \rangle$.


Proof

\(\displaystyle x \in S\) \(\leadsto\) \(\displaystyle x \in S \lor x \in T\) $\quad$ Rule of Addition $\quad$
\(\displaystyle \) \(\leadsto\) \(\displaystyle x \in S \cup T\) $\quad$ Definition of Set Union $\quad$
\(\displaystyle \) \(\leadsto\) \(\displaystyle S \subseteq S \cup T\) $\quad$ Definition of Subset $\quad$

Similarly for $T$.

$\blacksquare$


Sources