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:
\(\ds S_1 \cup S_2\) | \(\subseteq\) | \(\ds T \cup T\) | Set Union Preserves Subsets | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds S_1 \cup S_2\) | \(\subseteq\) | \(\ds T\) | Set 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.
$\blacksquare$
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 $1.2.1 \ \text{(iii)}$