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

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

Let $\mathcal P \left({T}\right)$ be the power set of $T$.

Let $\mathbb T \subseteq \mathcal P \left({T}\right)$.

Then:
 * $(1): \quad \displaystyle S \setminus \bigcap \mathbb T = \bigcup_{T\ ' \in \mathbb T} \left({S \setminus T\ '}\right)$


 * $(2): \quad \displaystyle S \setminus \bigcup \mathbb T = \bigcap_{T\ ' \in \mathbb T} \left({S \setminus T\ '}\right)$

where:
 * $\displaystyle \bigcap \mathbb T := \left\{{x: \forall T\ ' \in \mathbb T: x \in T\ '}\right\}$

i.e. the intersection of $\mathbb T$


 * $\displaystyle \bigcup \mathbb T := \left\{{x: \exists T\ ' \in \mathbb T: x \in T\ '}\right\}$

i.e. the union of $\mathbb T$.

First result
$(1): \quad \displaystyle S \setminus \bigcap \mathbb T = \bigcup_{T\ ' \in \mathbb T} \left({S \setminus T\ '}\right)$

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

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

Then:

Therefore:
 * $S \setminus \bigcap \mathbb T = \bigcup_{T\ ' \in \mathbb T} \left({S \setminus T\ '}\right)$

Second result
To prove that: $(2): \quad \displaystyle S \setminus \bigcup \mathbb T = \bigcap_{T\ ' \in \mathbb T} \left({S \setminus T\ '}\right)$

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

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

Then:

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

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.

Proof by Induction
Let $\mathbb T = \left\{{T_i: i \in I}\right\}$, where each $T_i$ is a set and $I$ is some finite indexing set.

Then:


 * $\displaystyle S \setminus \bigcap_{i \in I} T_i = \bigcup_{i \in I} \left({S \setminus T_i}\right)$


 * $\displaystyle S \setminus \bigcup_{i \in I} T_i = \bigcap_{i \in I} \left({S \setminus T_i}\right)$