Set Union Preserves Subsets

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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 $\left \langle {A_\alpha} \right \rangle_{\alpha \mathop \in I}$ and $\left \langle {B_\alpha} \right \rangle_{\alpha \mathop \in I}$ be indexed families of subsets of a set $S$.

Let:

$\forall \beta \in I: A_\beta \subseteq B_\beta$


Then:

$\displaystyle \bigcup_{\alpha \mathop \in I} A_\alpha \subseteq \bigcup_{\alpha \mathop \in I} B_\alpha$


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:

\(\displaystyle x \in A\) \(\implies\) \(\displaystyle x \in B\) $\quad$ Definition of Subset $\quad$
\(\displaystyle x \in S\) \(\implies\) \(\displaystyle x \in T\) $\quad$ Definition of Subset $\quad$


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:

$\left({x \in A \implies x \in B, \ x \in S \implies x \in T}\right) \implies \left({x \in A \lor x \in S \implies x \in B \lor x \in T}\right)$

The result follows directly from the definition of set union:

$\left({x \in A \implies x \in B, \ x \in S \implies x \in T}\right) \implies \left({x \in A \cup S \implies x \in B \cup T}\right)$

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.

$\blacksquare$


Also see