De Morgan's Laws (Set Theory)/Proof by Induction

Theorem
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:


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

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 propositions:


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

as:


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

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

First result
For all $n \in \N^*$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle S \setminus \bigcap_{i = 1}^n T_i = \bigcup_{i = 1}^n \left({S \setminus T_i}\right)$


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

First result: 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.

First result: 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:


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

First result: Induction Step
Now we need to show:


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

This is our induction step:

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

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

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

Second result
For all $n \in \N^*$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle S \setminus \bigcup_{i = 1}^n T_i = \bigcap_{i = 1}^n \left({S \setminus T_i}\right)$.


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

Second result: Base Case

 * $P(2)$ is the case:
 * $S \setminus \left({T_1 \cup T_2}\right) = \left({S \setminus T_1}\right) \cap \left({S \setminus T_2}\right)$

which has been proved.

This is our basis for the induction.

Second result: 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:


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

Second result: Induction Step
Now we need to show:


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

This is our induction step:

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

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

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