# De Morgan's Laws (Set Theory)/Proof by Induction/Difference with Intersection

## Theorem

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

Then:

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

## Proof

Let the cardinality $\left|{I}\right|$ 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:

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

as:

$\displaystyle S \setminus \bigcap_{i \mathop = 1}^n T_i = \bigcup_{i \mathop = 1}^n \left({S \setminus T_i}\right)$

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

For all $n \in \N_{>0}$, let $P \left({n}\right)$ be the proposition:

$\displaystyle S \setminus \bigcap_{i \mathop = 1}^n T_i = \bigcup_{i \mathop = 1}^n \left({S \setminus T_i}\right)$

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

### Base Case

$P(2)$ is the case:

$S \setminus \left({T_1 \cap T_2}\right) = \left({S \setminus T_1}\right) \cup \left({S \setminus T_2}\right)$

which has been proved.

This is our basis for the induction.

### Induction Hypothesis

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

So this is our induction hypothesis:

$\displaystyle S \setminus \bigcap_{i \mathop = 1}^k T_i = \bigcup_{i \mathop = 1}^k \left({S \setminus T_i}\right)$

### Induction Step

Now we need to show:

$\displaystyle S \setminus \bigcap_{i \mathop = 1}^{k+1} T_i = \bigcup_{i \mathop = 1}^{k+1} \left({S \setminus T_i}\right)$

This is our induction step:

 $\displaystyle S \setminus \bigcap_{i \mathop = 1}^{k+1} T_i$ $=$ $\displaystyle S \setminus \left({\bigcap_{i \mathop = 1}^k T_i \cap T_{k+1} }\right)$ Intersection is Associative $\displaystyle$ $=$ $\displaystyle \left({S \setminus \bigcap_{i \mathop = 1}^k T_i}\right) \cup \left({S \setminus T_{k+1} }\right)$ Base case $\displaystyle$ $=$ $\displaystyle \left({\bigcup_{i \mathop = 1}^k \left({S \setminus T_i}\right)}\right) \cup \left({S \setminus T_{k+1} }\right)$ Induction hypothesis $\displaystyle$ $=$ $\displaystyle \bigcup_{i \mathop = 1}^{k+1} \left({S \setminus T_i}\right)$ Union is Associative

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

Therefore:

$\displaystyle S \setminus \bigcap_{i \mathop = 1}^n T_i = \bigcup_{i \mathop = 1}^n \left({S \setminus T_i}\right)$

that is:

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

$\blacksquare$

## Source of Name

This entry was named for Augustus De Morgan.