# Set Union Preserves Subsets/Corollary

## 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 1

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.

$\blacksquare$

## Proof 2

Let $A$, $B$, and $S$ be sets.

Let $A \subseteq B$.

Let $x \in A \cup S$.

By the definition of union:

$x \in A$ or $x \in S$

Suppose $x \in A$.

Then by the definition of subset:

$x \in B$

Thus by the definition of union:

$x \in B \cup S$

Suppose instead that $x \in S$.

Then by the definition of union:

$x \in B \cup S$

Thus for all $x \in A \cup S$:

$x \in B \cup S$

The second result follows from Union is Commutative.

$\blacksquare$