Set Union Preserves Subsets/Corollary
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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.
$\blacksquare$
Sources
- 1971: Robert H. Kasriel: Undergraduate Topology ... (previous) ... (next): $\S 1.6$: Set Identities and Other Set Relations: Exercise $1 \ \text{(d)}$