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

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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:

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


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 propositions:

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

as:

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

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:

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


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


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:

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


First result: 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:

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

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

i.e.

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

$\blacksquare$


Second result

For all $n \in \N^*$, let $P \left({n}\right)$ be the proposition:

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


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


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:

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


Second result: Induction Step

Now we need to show:

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


This is our induction step:

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

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


Therefore:

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

i.e.:

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

$\blacksquare$


Source of Name

This entry was named for Augustus De Morgan.