Union with Superset is Superset

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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.

$\blacksquare$


Also see


Sources