Union of Subsets is Subset

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:

$\ds \bigcup \mathbb S \subseteq T$

Subset of Power Set

Let $S$ and $T$ be sets.

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

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

Then:

$\ds \paren {\forall X \in \mathbb S: X \subseteq T} \implies \bigcup \mathbb S \subseteq T$

Family of Sets

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

Then for all sets $X$:

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

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

Proof 1

Let:

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

Then:

$\blacksquare$

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.

$\blacksquare$

Sources

• 1967: George McCarty: Topology: An Introduction with Application to Topological Groups ... (previous) ... (next): Chapter $\text{I}$: Sets and Functions: Exercise $\text{B vii}$
• 1971: Robert H. Kasriel: Undergraduate Topology ... (previous) ... (next): $\S 1.6$: Set Identities and Other Set Relations: Exercise $3$
• 1993: Keith Devlin: The Joy of Sets: Fundamentals of Contemporary Set Theory (2nd ed.) ... (previous) ... (next): $\S 1$: Naive Set Theory: $\S 1.2$: Operations on Sets: Exercise $1.2.1 \ \text{(iii)}$