Union of Subsets is Subset

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Theorem

Let $S_1$, $S_2$, and $T$ be sets.

Let $S_1$ and $S_2$ both be subsets of $T$.


Then:

$S_1 \cup S_2 \subseteq T$


That is:

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


Set of Sets

Let $T$ be a set.

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

Suppose that for each $S \in \mathbb S$, $S \subseteq T$.


Then:

$\displaystyle \bigcup \mathbb S \subseteq T$


Family of Sets

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


Then for all sets $X$:

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

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


Proof 1

Let:

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


Then:

\(\displaystyle S_1 \cup S_2\) \(\subseteq\) \(\displaystyle T \cup T\) Set Union Preserves Subsets
\(\displaystyle \leadsto \ \ \) \(\displaystyle S_1 \cup S_2\) \(\subseteq\) \(\displaystyle T\) Union is Idempotent

$\blacksquare$


Proof 2

Let $x \in S_1 \cup S_2$.

By the definition of union, either $x \in S_1$ or $x \in S_2$.

By hypothesis, $S_1 \subseteq T$ and $S_2 \subseteq T$.

By definition of subset:

$x \in S_1 \implies x \in T$
$x \in S_2 \implies x \in T$

By Proof by Cases it follows that $x \in T$.

Hence the result by definition of subset.

$\blacksquare$


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