De Morgan's Laws (Set Theory)/Set Difference/General Case/Difference with Union

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Let $S$ and $T$ be sets.

Let $\mathcal P \left({T}\right)$ be the power set of $T$.

Let $\mathbb T \subseteq \mathcal P \left({T}\right)$.


$\displaystyle S \setminus \bigcup \mathbb T = \bigcap_{T' \mathop \in \mathbb T} \left({S \setminus T'}\right)$


$\displaystyle \bigcup \mathbb T := \left\{{x: \exists T' \in \mathbb T: x \in T'}\right\}$

that is, the union of $\mathbb T$.



$\displaystyle x \in S \setminus \bigcup \mathbb T$

Note that by Set Difference is Subset we have that $x \in S$ (we need this later).


\(\displaystyle x\) \(\in\) \(\displaystyle S \setminus \bigcup \mathbb T\)
\(\displaystyle \iff \ \ \) \(\displaystyle x\) \(\notin\) \(\displaystyle \bigcup \mathbb T\) Definition of Set Difference
\(\displaystyle \iff \ \ \) \(\displaystyle \neg (\exists T' \in \mathbb T: (x\) \(\in\) \(\displaystyle T'))\) Definition of Set Union
\(\displaystyle \iff \ \ \) \(\displaystyle \forall T' \in \mathbb T: \neg (x\) \(\in\) \(\displaystyle T')\) De Morgan's Laws (Predicate Logic)
\(\displaystyle \iff \ \ \) \(\displaystyle \forall T' \in \mathbb T: x\) \(\in\) \(\displaystyle S \setminus T'\) Definition of Set Difference: note $x \in S$ from above
\(\displaystyle \iff \ \ \) \(\displaystyle x\) \(\in\) \(\displaystyle \bigcap_{T' \mathop \in \mathbb T} \left({S \setminus T'}\right)\) Definition of Set Intersection


$\displaystyle S \setminus \mathbb T = \bigcap_{T' \mathop \in \mathbb T} \left({S \setminus T'}\right)$



It is tempting to set up an argument to prove the general case using induction. While this works, and is a perfectly valid demonstration for an elementary student in how such proofs are crafted, such a proof is inadequate as it is valid only when $\mathbb T$ is finite.

The proof as given above relies only upon De Morgan's laws as applied to predicate logic. Thus the uncountable case has been reduced to a result in logic.

However, for better or worse, the following is an example of how one might achieve this result using induction.

Proof by Induction

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:

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


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

The proof of 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:

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

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

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.

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:

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

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:

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

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


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

that is:

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


Source of Name

This entry was named for Augustus De Morgan.