Union of Subsets is Subset/Proof 2

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:

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

Proof

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

• 1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 1.4$. Union: Example $15$
• 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 7.2 \ \text{(ii)}$: Unions and Intersections