De Morgan's Laws (Set Theory)/Set Difference/Family of Sets

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

Let $\left\langle{T_i}\right\rangle_{i \in I}$ be a family of subsets of $T$.

Then:
 * $(1): \quad \displaystyle S \setminus \bigcap_{i \mathop \in I} T_i = \bigcup_{i \mathop \in I} \left({S \setminus T_i}\right)$


 * $(2): \quad \displaystyle S \setminus \bigcup_{i \mathop \in I} T_i = \bigcap_{i \mathop \in I} \left({S \setminus T_i}\right)$

where:
 * $\displaystyle \bigcap_{i \mathop \in I} T_i := \left\{{x: \forall i \in I: x \in T_i}\right\}$

i.e. the intersection of $\left\langle{T_i}\right\rangle_{i \in I}$


 * $\displaystyle \bigcup_{i \mathop \in I} T_i := \left\{{x: \exists i \in I: x \in T_i}\right\}$

i.e. the union of $\left\langle{T_i}\right\rangle_{i \in I}$.

First result
$(1): \quad \displaystyle S \setminus \bigcap_{i \mathop \in I} T_i = \bigcup_{i \mathop \in I} \left({S \setminus T_i}\right)$

Suppose:
 * $\displaystyle x \in S \setminus \bigcap_{i \mathop \in I} T_i$

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

Then:

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

Second result
To prove that: $(2): \quad \displaystyle S \setminus \bigcup_{i \mathop \in I} T_i = \bigcap_{i \mathop \in I} \left({S \setminus T_i}\right)$

Suppose:
 * $\displaystyle x \in S \setminus \bigcup_{i \mathop \in I} T_i$

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

Then:

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