# 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:

$\ds S \setminus \bigcup \mathbb T = \bigcap_{T' \mathop \in \mathbb T} \paren {S \setminus T'}$

where:

$\ds \bigcup \mathbb T := \set {x: \exists T' \in \mathbb T: x \in T'}$

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

## Proof

Suppose:

$\ds x \in S \setminus \bigcup \mathbb T$

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

Then:

 $\ds x$ $\in$ $\ds S \setminus \bigcup \mathbb T$ $\ds \leadstoandfrom \ \$ $\ds x$ $\notin$ $\ds \bigcup \mathbb T$ Definition of Set Difference $\ds \leadstoandfrom \ \$ $\ds \nexists T' \in \mathbb T: \,$ $\ds x$ $\in$ $\ds T'$ Definition of Union of Set of Sets $\ds \leadstoandfrom \ \$ $\ds \forall T' \in \mathbb T: \,$ $\ds x$ $\notin$ $\ds T'$ Denial of Existence $\ds \leadstoandfrom \ \$ $\ds \forall T' \in \mathbb T: \,$ $\ds x$ $\in$ $\ds S \setminus T'$ Definition of Set Difference: note $x \in S$ from above $\ds \leadstoandfrom \ \$ $\ds x$ $\in$ $\ds \bigcap_{T' \mathop \in \mathbb T} \paren {S \setminus T'}$ Definition of Intersection of Set of Sets

Therefore:

$\ds S \setminus \mathbb T = \bigcap_{T' \mathop \in \mathbb T} \paren {S \setminus T'}$

$\blacksquare$

## 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 the cardinality $\size I$ of the indexing set $I$ be $n$.

Then by the definition of cardinality, it follows that $I \cong \N^*_n$ and we can express the proposition:

$\ds S \setminus \bigcup_{i \mathop \in I} T_i = \bigcap_{i \mathop \in I} \paren {S \setminus T_i}$

as:

$\ds S \setminus \bigcup_{i \mathop = 1}^n T_i = \bigcap_{i \mathop = 1}^n \paren {S \setminus T_i}$

The proof of this is more amenable to proof by Principle of Mathematical Induction.

For all $n \in \N_{>0}$, let $\map P n$ be the proposition:

$\ds S \setminus \bigcup_{i \mathop = 1}^n T_i = \bigcap_{i \mathop = 1}^n \paren {S \setminus T_i}$.

$\map P 1$ is true, as this just says $S \setminus T_1 = S \setminus T_1$.

### Base Case

$\map P 2$ is the case:

$S \setminus \paren {T_1 \cup T_2} = \paren {S \setminus T_1} \cap \paren {S \setminus T_2}$

which has been proved.

This is our basis for the induction.

### Induction Hypothesis

Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:

$\ds S \setminus \bigcup_{i \mathop = 1}^k T_i = \bigcap_{i \mathop = 1}^k \paren {S \setminus T_i}$

### Induction Step

Now we need to show:

$\ds S \setminus \bigcup_{i \mathop = 1}^{k + 1} T_i = \bigcap_{i \mathop = 1}^{k + 1} \paren {S \setminus T_i}$

This is our induction step:

 $\ds S \setminus \bigcup_{i \mathop = 1}^{k + 1} T_i$ $=$ $\ds S \setminus \paren {\bigcup_{i \mathop = 1}^k T_i \cup T_{k + 1} }$ Union is Associative $\ds$ $=$ $\ds \paren {S \setminus \bigcup_{i \mathop = 1}^k T_i} \cap \paren {S \setminus T_{k + 1} }$ Base case $\ds$ $=$ $\ds \paren {\bigcap_{i \mathop = 1}^k \paren {S \setminus T_i} } \cap \paren {S \setminus T_{k + 1} }$ Induction hypothesis $\ds$ $=$ $\ds \bigcap_{i \mathop = 1}^{k + 1} \paren {S \setminus T_i}$ Intersection is Associative

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:

$\ds S \setminus \bigcup_{i \mathop = 1}^n T_i = \bigcap_{i \mathop = 1}^n \paren {S \setminus T_i}$

that is:

$\ds S \setminus \bigcup_{i \mathop \in I} T_i = \bigcap_{i \mathop \in I} \paren {S \setminus T_i}$

$\blacksquare$

## Source of Name

This entry was named for Augustus De Morgan.