# Union with Superset is Superset

## Theorem

$S \subseteq T \iff S \cup T = T$

where:

$S \subseteq T$ denotes that $S$ is a subset of $T$
$S \cup T$ denotes the union of $S$ and $T$.

## Proof 1

Let $S \cup T = T$.

Then by definition of set equality:

$S \cup T \subseteq T$

Thus:

 $\ds S$ $\subseteq$ $\ds S \cup T$ Subset of Union $\ds \leadsto \ \$ $\ds S$ $\subseteq$ $\ds T$ Subset Relation is Transitive

Now let $S \subseteq T$.

From Subset of Union, we have $S \cup T \supseteq T$.

We also have:

 $\ds S$ $\subseteq$ $\ds T$ $\ds \leadsto \ \$ $\ds S \cup T$ $\subseteq$ $\ds T \cup T$ Set Union Preserves Subsets $\ds \leadsto \ \$ $\ds S \cup T$ $\subseteq$ $\ds T$ Set Union is Idempotent

Then:

 $\ds S \cup T$ $\subseteq$ $\ds T$ $\ds S \cup T$ $\supseteq$ $\ds T$

By definition of set equality:

$S \cup T = T$

So:

 $\ds S \cup T = T$ $\implies$ $\ds S \subseteq T$ $\ds S \subseteq T$ $\implies$ $\ds S \cup T = T$

and so:

$S \subseteq T \iff S \cup T = T$

from the definition of equivalence.

$\blacksquare$

## Proof 2

 $\ds$  $\ds S \cup T = T$ $\ds \leadstoandfrom \ \$ $\ds$  $\ds \paren {x \in S \lor x \in T \iff x \in T}$ Definition of Set Equality $\ds \leadstoandfrom \ \$ $\ds$  $\ds \paren {x \in S \implies x \in T}$ Conditional iff Biconditional of Consequent with Disjunction $\ds \leadstoandfrom \ \$ $\ds$  $\ds S \subseteq T$ Definition of Subset

$\blacksquare$