Union is Smallest Superset

Theorem
$$\left({R \subseteq T}\right) \land \left({S \subseteq T}\right) \iff \left({R \cup S}\right) \subseteq T$$

Thus $$R \cup S$$ is the smallest subset of $$T$$ containing both $$R$$ and $$S$$, in the sense that for any $$A \subseteq T$$, if $$R \subseteq A$$ and $$S \subseteq A$$ then $$\left({R \cup S}\right) \subseteq A$$.

Proof

 * First we show $$\left({R \subseteq T}\right) \land \left({S \subseteq T}\right) \Longrightarrow \left({R \cup S}\right) \subseteq T$$:

$$\left({R \subseteq T}\right) \land \left({S \subseteq T}\right)$$

$$\Longrightarrow \left({x \in R \Longrightarrow x \in T}\right) \land \left({x \in S \Longrightarrow x \in T}\right)$$ Subset definition

$$x \in \left({R \cup S}\right)$$

$$\Longrightarrow \left({x \in R}\right) \lor \left({x \cup S}\right)$$ Definition of Set Union

$$\Longrightarrow \left({x \in T}\right) \lor \left({x \in T}\right)$$ Constructive Dilemma

$$\Longrightarrow x \in T$$ Rule of Idempotence

$$\Longrightarrow \left({R \cup S}\right) \subseteq T$$


 * Next we show $$\left({R \cup S}\right) \subseteq T \Longrightarrow \left({R \subseteq T}\right) \land \left({S \subseteq T}\right)$$:

$$R \subset R \cup S$$ Subset of Union

$$\left({R \cup S}\right) \subseteq T$$ by hypothesis

$$\Longrightarrow R \subseteq T$$ Subsets Transitive

Similarly for $$S$$:

$$S \subset R \cup S$$ Subset of Union

$$\left({R \cup S}\right) \subseteq T$$ by hypothesis

$$\Longrightarrow S \subseteq T$$ Subsets Transitive