# 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

Let $S \cup T = T$.

Then by definition of set equality:

$S \cup T \subseteq T$

Thus:

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

Now let $S \subseteq T$.

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

We also have:

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

Then:

 $\displaystyle S \cup T$ $\subseteq$ $\displaystyle T$ $\displaystyle S \cup T$ $\supseteq$ $\displaystyle T$

By definition of set equality:

$S \cup T = T$

So:

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

and so:

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

from the definition of equivalence.

$\blacksquare$