De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Union

Theorem
Let $S$ and $T$ be sets.

Let $\powerset T$ be the power set of $T$.

Let $\mathbb T \subseteq \powerset T$.

Then:
 * $\displaystyle S \setminus \bigcup \mathbb T = \bigcap_{T' \mathop \in \mathbb T} \paren {S \setminus T'}$

where:
 * $\displaystyle \bigcup \mathbb T := \set {x: \exists T' \in \mathbb T: x \in T'}$

that is, the union of $\mathbb T$.

Proof
Suppose:
 * $\displaystyle x \in S \setminus \bigcup \mathbb T$

Note that by Set Difference is Subset we have that $x \in S$ (we need this later).

Then:

Therefore:
 * $\displaystyle S \setminus \mathbb T = \bigcap_{T' \mathop \in \mathbb T} \left({S \setminus T'}\right)$

Caution
It is tempting to set up an argument to prove the general case using induction. While this works, and is a perfectly valid demonstration for an elementary student in how such proofs are crafted, such a proof is inadequate as it is valid only when $\mathbb T$ is finite.

The proof as given above relies only upon De Morgan's laws as applied to predicate logic. Thus the uncountable case has been reduced to a result in logic.

However, for better or worse, the following is an example of how one might achieve this result using induction.