Set Union Preserves Subsets/Corollary/Proof 1

Corollary to Set Union Preserves Subsets

Let $A, B, S$ be sets.

Then:

$A \subseteq B \implies A \cup S \subseteq B \cup S$

$A \subseteq B \implies S \cup A \subseteq S \cup B$

Proof

Let $A \subseteq B$, and let $S$ be any set.

From Set Union Preserves Subsets, substituting $S$ for $T$:

$A \subseteq B, \ S \subseteq S \implies A \cup S \subseteq B \cup S$

From Set is Subset of Itself, $S \subseteq S$ for all sets $S$.

Hence the first result:

$A \subseteq B \implies A \cup S \subseteq B \cup S$

The second result follows from Union is Commutative.

$\blacksquare$

Sources

• 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: $\text{(e)}$