Union Smallest

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:

$\left({S_1 \subseteq T}\right) \land \left({S_2 \subseteq T}\right) \iff \left({S_1 \cup S_2}\right) \subseteq T$

General Result

Let $S$ be a set.

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

Let $\mathbb S$ be a subset of $\mathcal P \left({S}\right)$.

Then:

$\displaystyle \left({\forall X \in \mathbb S: X \subseteq T}\right) \iff \bigcup \mathbb S \subseteq T$

Proof

• Let $\left({S_1 \subseteq T}\right) \land \left({S_2 \subseteq T}\right)$.

Then:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle S_1\)	\(\subseteq\)	\(\displaystyle T\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle S_1 \cup S_2\)	\(\subseteq\)	\(\displaystyle T \cup S_2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Set Union Preserves Subsets
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle S_1 \cup S_2\)	\(\subseteq\)	\(\displaystyle T\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	as $T = T \cup S_2$ from Union with Superset is Superset‎

So:

$\left({S_1 \subseteq T}\right) \land \left({S_2 \subseteq T}\right) \implies \left({S_1 \cup S_2}\right) \subseteq T$

• Next we show $\left({S_1 \cup S_2}\right) \subseteq T \implies \left({S_1 \subseteq T}\right) \land \left({S_2 \subseteq T}\right)$:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle S_1\)	\(\subseteq\)	\(\displaystyle S_1 \cup S_2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Subset of Union
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left({S_1 \cup S_2}\right)\)	\(\subseteq\)	\(\displaystyle T\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by hypothesis
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle S_1\)	\(\subseteq\)	\(\displaystyle T\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Subsets Transitive

Similarly for $S$:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle S_2\)	\(\subseteq\)	\(\displaystyle S_1 \cup S_2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Subset of Union
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left({S_1 \cup S_2}\right)\)	\(\subseteq\)	\(\displaystyle T\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by hypothesis
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle S_2\)	\(\subseteq\)	\(\displaystyle T\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Subsets Transitive

• So, from the above, we have:

1. $\left({S_1 \subseteq T}\right) \land \left({S_2 \subseteq T}\right) \implies \left({S_1 \cup S_2}\right) \subseteq T$
2. $\left({S_1 \cup S_2}\right) \subseteq T \implies \left({S_1 \subseteq T}\right) \land \left({S_2 \subseteq T}\right)$

Thus $\left({S_1 \subseteq T}\right) \land \left({S_2 \subseteq T}\right) \iff \left({S_1 \cup S_2}\right) \subseteq T$ from the definition of equivalence.

$\blacksquare$

Proof of General Result

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

Suppose that $\forall X \in \mathbb S: X \subseteq T$.

Consider any $\displaystyle x \in \bigcup \mathbb S$.

By definition of set union, it follows that:

$\exists X \in \mathbb S: x \in X$

But as $X \subseteq T$ it follows that $x \in T$.

Thus it follows that:

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

So:

$\displaystyle \left({\forall X \in \mathbb S: X \subseteq T}\right) \implies \bigcup \mathbb S \subseteq T$

$\Box$

Now suppose that $\displaystyle \bigcup \mathbb S \subseteq T$.

Consider any $X \in \mathbb S$ and take any $x \in X$.

From Subset of Union: General Result we have that $X \subseteq \bigcup \mathbb S$.

Thus $\displaystyle x \in \bigcup \mathbb S$.

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

So it follows that $X \subseteq T$.

So:

$\displaystyle \bigcup \mathbb S \subseteq T \implies \left({\forall X \in \mathbb S: X \subseteq T}\right)$

$\Box$

Hence:

$\displaystyle \left({\forall X \in \mathbb S: X \subseteq T}\right) \iff \bigcup \mathbb S \subseteq T$

$\blacksquare$

Also see

• Intersection Largest

Sources

• J.A. Green: Sets and Groups (1965)... (previous)... (next): $\S 1.4$: Example $15$
• Seth Warner: Modern Algebra (1965)... (previous)... (next): Exercise $3.6 \ \text{(f)}$
• George McCarty: Topology: An Introduction with Application to Topological Groups (1967): $\text{I}$: Exercises $\text{B vii}$
• T.S. Blyth: Set Theory and Abstract Algebra (1975): $\S 1, \S 6$, Theorem $6.1 \ (2)$
• Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): $\S 7.2 \ \text{(ii)}$
• Keith Devlin: The Joy of Sets: Fundamentals of Contemporary Set Theory (1993): $\S 1.2$: Exercise $1.2.1 \ \text{(iii)}$
• Keith Devlin: The Joy of Sets: Fundamentals of Contemporary Set Theory (1993): $\S 1.4$: Exercise $1.4.4 \ \text{(i)}$