Subset of Union
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Contents |
Theorem
The union of two sets is a superset of each:
- $S \subseteq S \cup T$
- $T \subseteq S \cup T$
General Result
Let $S$ be a set.
Let $\mathcal P \left({S}\right)$ be the power set of $S$.
Let $\mathbb S \subseteq \mathcal P \left({S}\right)$.
Then:
- $\forall T \in \mathbb S: T \subseteq \bigcup \mathbb S$
Proof
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle x \in S\) | \(\implies\) | \(\displaystyle x \in S \lor x \in T\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Rule of Addition | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\implies\) | \(\displaystyle x \in S \cup T\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Definition of Set Union | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\implies\) | \(\displaystyle S \subseteq S \cup T\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Definition of Subset |
Similarly for $T$.
$\blacksquare$
Proof of General Result
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \forall T \in \mathbb S: x \in T\) | \(\implies\) | \(\displaystyle x \in \bigcup \mathbb S\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Definition of Set Union | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\implies\) | \(\displaystyle \forall T \in \mathbb S: T \subseteq \bigcup \mathbb S\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Definition of Subset |
$\blacksquare$
Sources
- W.E. Deskins: Abstract Algebra (1964): $\S 1.1$: Theorem $1.4$, Exercise $1.1: 8 \ \text{(f)}$
- J.A. Green: Sets and Groups (1965)... (previous)... (next): $\S 1.4$: Example $15$
- Allan Clark: Elements of Abstract Algebra (1971)... (previous)... (next): $\S 6$
- Keith Devlin: The Joy of Sets: Fundamentals of Contemporary Set Theory (1993): $\S 1.2$: Exercise $1.2.1 \ \text{(ii)}$