Subset Equivalences

Theorem
These statements concerning subsets are equivalent:


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

where:
 * $$S \subseteq T$$ denotes that $$S$$ is a subset of $$T$$;
 * $$S \cup T$$ denotes the union of $$S$$ and $$T$$;
 * $$S \cap T$$ denotes the intersection of $$S$$ and $$T$$;
 * $$S - T$$ denotes the set difference between $$S$$ and $$T$$;
 * $$\varnothing$$ denotes the empty set;
 * $$\mathcal{C}$$ denotes the complement of $$S$$;
 * $$\mathbb{U}$$ denotes the universal set.

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

Then from the definition of set equality, $$S \cup T \subseteq T$$.

Thus:

$$ $$

Now let $$S \subseteq T$$. Then:

$$ $$ $$

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

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

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

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

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

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

Then by the definition of set equality, $$S \subseteq S \cap T$$.

Thus:

$$ $$

Now let $$S \subseteq T$$.

We have:

$$ $$ $$

Then we have:

$$ $$

So we have $$S \cap T = S \implies S \subseteq T$$ and $$S \subseteq T \implies S \cap T = S$$ 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 $$S \subseteq T \iff \mathcal{C} \left({T}\right) \subseteq \mathcal{C} \left({S}\right)$$.