Multiplicatively Closed Subset is Saturated iff Complement is Union of Prime Ideals

Definition
Let $A$ be a commutative ring with unity.

Let $S \subseteq A$ be a multiplicatively closed subset.

Then the following are equivalent:
 * $(1):\quad$ $S$ is saturated
 * $(2):\quad$ $\relcomp A S = \bigcup \set {\mathfrak p \in \Spec A : \mathfrak p \cap S = \O }$
 * $(3):\quad$ $\exists \TT \subseteq \Spec A : \relcomp A S = \bigcup \TT$

where:
 * $\relcomp A S$ denotes the complement of $S$
 * $\Spec A$ denotes the prime spectrum of $A$

$(2) \implies (3)$
This is clear, choosing:
 * $\TT = \set {\mathfrak p \in \Spec A : \mathfrak p \cap S = \O }$.

$(3) \implies (1)$
Let $\TT \subseteq \Spec A$ be such that:
 * $\relcomp A S = \bigcup \TT$.

Let $x, y \in A$ be such that $x y \in S$.

That is:
 * $\forall \mathfrak p \in \TT : xy \not \in \mathfrak p$

As $\mathfrak p$s are ideals, we have:
 * $\forall \mathfrak p \in \TT : x \not \in \mathfrak p \land y \not \in \mathfrak p$

That is:
 * $x, y \in S$