Set Union Preserves Subsets
Theorem
Let $A, B, S, T$ be sets.
Then:
- $A \subseteq B, \ S \subseteq T \implies A \cup S \subseteq B \cup T$
Corollary
Let $A, B, S$ be sets.
Then:
- $A \subseteq B \implies A \cup S \subseteq B \cup S$
- $A \subseteq B \implies S \cup A \subseteq S \cup B$
Families of Sets
Let $I$ be an indexing set.
Let $\family {A_\alpha}_{\alpha \mathop \in I}$ and $\family {B_\alpha}_{\alpha \mathop \in I}$ be indexed families of subsets of a set $S$.
Let:
- $\forall \beta \in I: A_\beta \subseteq B_\beta$
Then:
- $\ds \bigcup_{\alpha \mathop \in I} A_\alpha \subseteq \bigcup_{\alpha \mathop \in I} B_\alpha$
General Result
Let $\mathbb S, \mathbb T$ be sets of sets.
Suppose that for each $S \in \mathbb S$ there exists a $T \in \mathbb T$ such that $S \subseteq T$.
Then $\bigcup \mathbb S \subseteq \bigcup \mathbb T$.
Proof 1
Let $A \subseteq B$ and $S \subseteq T$.
Then:
| \(\ds x \in A\) | \(\leadsto\) | \(\ds x \in B\) | Definition of Subset | |||||||||||
| \(\ds x \in S\) | \(\leadsto\) | \(\ds x \in T\) | Definition of Subset |
Now we invoke the Constructive Dilemma of propositional logic:
- $p \implies q, \ r \implies s \vdash p \lor r \implies q \lor s$
applying it as:
- $\paren {x \in A \implies x \in B, \ x \in S \implies x \in T} \implies \paren {x \in A \lor x \in S \implies x \in B \lor x \in T}$
The result follows directly from the definition of set union:
- $\paren {x \in A \implies x \in B, \ x \in S \implies x \in T} \implies \paren {x \in A \cup S \implies x \in B \cup T}$
and from the definition of subset:
- $A \subseteq B, \ S \subseteq T \implies A \cup S \subseteq B \cup T$
$\blacksquare$
Proof 2
By Subset Relation is Transitive, $\subseteq$ is a transitive relation.
By the corollary to Set Union Preserves Subsets (Proof 2), $\subseteq$ is compatible with $\cup$.
Thus the theorem holds by Operating on Transitive Relationships Compatible with Operation.
$\blacksquare$