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 indexing set.

Then:


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

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:


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

as:


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

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

First result

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

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


 * $$P(1)$$ 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:


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

First result: Induction Step
Now we need to show:


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

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 $$S \setminus \bigcap_{i = 1}^n T_i = \bigcup_{i = 1}^n \left({S \setminus T_i}\right)$$, i.e. $$S \setminus \bigcap_{i \in I} T_i = \bigcup_{i \in I} \left({S \setminus T_i}\right)$$.

Second result

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

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


 * $$P(1)$$ 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:


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

Second result: Induction Step
Now we need to show:


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

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 $$S \setminus \bigcup_{i = 1}^n T_i = \bigcap_{i = 1}^n \left({S \setminus T_i}\right)$$, i.e. $$S \setminus \bigcup_{i \in I} T_i = \bigcap_{i \in I} \left({S \setminus T_i}\right)$$.