# 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:

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

### 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:

 $\displaystyle x \in A$ $\leadsto$ $\displaystyle x \in B$ Definition of Subset $\displaystyle x \in S$ $\leadsto$ $\displaystyle x \in T$ 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.

$\blacksquare$