# Union is Smallest Superset

## Theorem

Let $S_1$ and $S_2$ be sets.

Then $S_1 \cup S_2$ is the smallest set containing both $S_1$ and $S_2$.

That is:

$\paren {S_1 \subseteq T} \land \paren {S_2 \subseteq T} \iff \paren {S_1 \cup S_2} \subseteq T$

### Set of Sets

Let $T$ be a set.

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

Then:

$\displaystyle \paren {\forall X \in \mathbb S: X \subseteq T} \iff \bigcup \mathbb S \subseteq T$

### General Result

Let $S$ and $T$ be sets.

Let $\powerset S$ denote the power set of $S$.

Let $\mathbb S$ be a subset of $\powerset S$.

Then:

$\displaystyle \paren {\forall X \in \mathbb S: X \subseteq T} \iff \bigcup \mathbb S \subseteq T$

### Family of Sets

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

Let $\family {S_i}_{i \mathop \in I}$ be a family of sets indexed by $I$.

Then for all sets $X$:

$\displaystyle \paren {\forall i \in I: S_i \subseteq X} \iff \bigcup_{i \mathop \in I} S_i \subseteq X$

where $\displaystyle \bigcup_{i \mathop \in I} S_i$ is the union of $\family {S_i}$.

## Proof

### Necessary Condition

$\paren {S_1 \subseteq T} \land \paren {S_2 \subseteq T} \implies \paren {S_1 \cup S_2} \subseteq T$

$\Box$

### Sufficient Condition

Let $\paren {S_1 \cup S_2} \subseteq T$:

 $\displaystyle S_1$ $\subseteq$ $\displaystyle S_1 \cup S_2$ Set is Subset of Union $\displaystyle \paren {S_1 \cup S_2}$ $\subseteq$ $\displaystyle T$ by hypothesis $\displaystyle \leadsto \ \$ $\displaystyle S_1$ $\subseteq$ $\displaystyle T$ Subset Relation is Transitive

Similarly for $S_2$:

 $\displaystyle S_2$ $\subseteq$ $\displaystyle S_1 \cup S_2$ Set is Subset of Union $\displaystyle \paren {S_1 \cup S_2}$ $\subseteq$ $\displaystyle T$ by hypothesis $\displaystyle \leadsto \ \$ $\displaystyle S_2$ $\subseteq$ $\displaystyle T$ Subset Relation is Transitive

That is:

$\paren {S_1 \cup S_2} \subseteq T \implies \paren {S_1 \subseteq T} \land \paren {S_2 \subseteq T}$

$\Box$

Thus from the definition of equivalence:

$\paren {S_1 \subseteq T} \land \paren {S_2 \subseteq T} \iff \paren {S_1 \cup S_2} \subseteq T$

$\blacksquare$