Subset Equivalences

Theorem
These statements are equivalent:


 * $$S \subseteq T$$
 * $$S \cup T = T$$
 * $$S \cap T = S$$
 * $$S - T = \varnothing$$
 * $$S \cap \mathcal{C} \left({T}\right) = \varnothing$$
 * $$\mathcal{C} \left({S}\right) \cup T = \mathbb{U}$$
 * $$\mathcal{C} \left({T}\right) \subseteq \mathcal{C} \left({S}\right)$$

Proof
Let $$S \cup T = T$$.

From Subset of Union, $$S \subseteq S \cup T$$

Hence from Subsets Transitive, $$S \subseteq T$$.

Thus $$S \cup T = T \Longrightarrow S \subseteq T$$.

Now let $$S \subseteq T$$.

From Subset of Itself, $$T \subseteq T$$.

By the Rule of Conjunction, $$S \subseteq T \land T \subseteq T$$.

Thus from Union Smallest, $$S \cup T \subseteq T$$

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

So we have $$T \subseteq S \cup T \land S \cup T \subseteq T$$.

Hence by the definition of set equality, $$S \cup T = T$$.

Thus $$S \subseteq T \Longrightarrow S \cup T = T$$.

Hence $$S \subseteq T \iff S \cup T = T$$, so $$S \cup T = T$$ and $$S \subseteq T$$ are equivalent.

Let $$S \cap T = S$$.

From Intersection Subset, $$S \cap T \subseteq T$$

So $$S \subseteq T$$.

Thus $$S \cap T = S \Longrightarrow S \subseteq T$$

Now let $$S \subseteq T$$.

We have $$S \subseteq S$$ Subset of Itself

$$\Longrightarrow S \subseteq S \land S \subseteq T$$ Rule of Conjunction

$$\Longrightarrow S \subseteq S \cap T$$ Intersection Largest

$$S \cap T \subseteq S$$ Intersection Subset

$$\Longrightarrow S \cap T \subseteq S$$ Set Equality

So $$S \subseteq T \Longrightarrow S \cap T = S$$ and

So we have $$S \cap T = S \Longrightarrow S \subseteq T$$ and $$S \subseteq T \Longrightarrow S \cap T = S$$ and so $$S \subseteq T \iff S \cap T = S$$.

Thus $$S \subseteq T \iff S - T = \varnothing$$.

Thus $$S \subseteq T \iff S \cap \mathcal{C} \left({T}\right) = \varnothing$$.

Thus $$S \subseteq T \iff \mathcal{C} \left({S}\right) \cup T = \mathbb{U}$$

Thus $$S \subseteq T \iff \mathcal{C} \left({T}\right) \subseteq \mathcal{C} \left({S}\right)$$.