De Morgan's Laws (Set Theory)/Proof by Induction/Difference with Intersection
Theorem
Let $\mathbb T = \set {T_i: i \mathop \in I}$, where each $T_i$ is a set and $I$ is some finite indexing set.
Then:
- $\ds S \setminus \bigcap_{i \mathop \in I} T_i = \bigcup_{i \mathop \in I} \paren {S \setminus T_i}$
Proof
Let the cardinality $\card I$ 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:
- $\ds S \setminus \bigcap_{i \mathop \in I} T_i = \bigcup_{i \mathop \in I} \paren {S \setminus T_i}$
as:
- $\ds S \setminus \bigcap_{i \mathop = 1}^n T_i = \bigcup_{i \mathop = 1}^n \paren {S \setminus T_i}$
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:
- $\ds S \setminus \bigcap_{i \mathop = 1}^n T_i = \bigcup_{i \mathop = 1}^n \paren {S \setminus T_i}$
$\map P 1$ is true, as this just says $S \setminus T_1 = S \setminus T_1$.
Base Case
$\map P 2$ is the case:
- $S \setminus \paren {T_1 \cap T_2} = \paren {S \setminus T_1} \cup \paren {S \setminus T_2}$
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:
- $\ds S \setminus \bigcap_{i \mathop = 1}^k T_i = \bigcup_{i \mathop = 1}^k \paren {S \setminus T_i}$
Induction Step
Now we need to show:
- $\ds S \setminus \bigcap_{i \mathop = 1}^{k + 1} T_i = \bigcup_{i \mathop = 1}^{k + 1} \paren {S \setminus T_i}$
This is our induction step:
\(\ds S \setminus \bigcap_{i \mathop = 1}^{k + 1} T_i\) | \(=\) | \(\ds S \setminus \paren {\bigcap_{i \mathop = 1}^k T_i \cap T_{k + 1} }\) | Intersection is Associative | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {S \setminus \bigcap_{i \mathop = 1}^k T_i} \cup \paren {S \setminus T_{k + 1} }\) | Base case | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\bigcup_{i \mathop = 1}^k \paren {S \setminus T_i} } \cup \paren {S \setminus T_{k + 1} }\) | Induction hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \bigcup_{i \mathop = 1}^{k + 1} \paren {S \setminus T_i}\) | Union is Associative |
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\ds S \setminus \bigcap_{i \mathop = 1}^n T_i = \bigcup_{i \mathop = 1}^n \paren {S \setminus T_i}$
that is:
- $\ds S \setminus \bigcap_{i \mathop \in I} T_i = \bigcup_{i \mathop \in I} \paren {S \setminus T_i}$
$\blacksquare$
Source of Name
This entry was named for Augustus De Morgan.