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

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Let $S$ and $T$ be sets.

Let $\powerset T$ be the power set of $T$.

Let $\mathbb T \subseteq \powerset T$.


$\ds S \setminus \bigcup \mathbb T = \bigcap_{T' \mathop \in \mathbb T} \paren {S \setminus T'}$


$\ds \bigcup \mathbb T := \set {x: \exists T' \in \mathbb T: x \in T'}$

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



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

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


\(\ds x\) \(\in\) \(\ds S \setminus \bigcup \mathbb T\)
\(\ds \leadstoandfrom \ \ \) \(\ds x\) \(\notin\) \(\ds \bigcup \mathbb T\) Definition of Set Difference
\(\ds \leadstoandfrom \ \ \) \(\ds \nexists T' \in \mathbb T: \, \) \(\ds x\) \(\in\) \(\ds T'\) Definition of Union of Set of Sets
\(\ds \leadstoandfrom \ \ \) \(\ds \forall T' \in \mathbb T: \, \) \(\ds x\) \(\notin\) \(\ds T'\) Denial of Existence
\(\ds \leadstoandfrom \ \ \) \(\ds \forall T' \in \mathbb T: \, \) \(\ds x\) \(\in\) \(\ds S \setminus T'\) Definition of Set Difference: note $x \in S$ from above
\(\ds \leadstoandfrom \ \ \) \(\ds x\) \(\in\) \(\ds \bigcap_{T' \mathop \in \mathbb T} \paren {S \setminus T'}\) Definition of Intersection of Set of Sets


$\ds S \setminus \mathbb T = \bigcap_{T' \mathop \in \mathbb T} \paren {S \setminus T'}$



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 $\size I$ 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:

$\ds S \setminus \bigcup_{i \mathop \in I} T_i = \bigcap_{i \mathop \in I} \paren {S \setminus T_i}$


$\ds S \setminus \bigcup_{i \mathop = 1}^n T_i = \bigcap_{i \mathop = 1}^n \paren {S \setminus T_i}$

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:

$\ds S \setminus \bigcup_{i \mathop = 1}^n T_i = \bigcap_{i \mathop = 1}^n \paren {S \setminus T_i}$.

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

Base Case

$\map P 2$ is the case:

$S \setminus \paren {T_1 \cup T_2} = \paren {S \setminus T_1} \cap \paren {S \setminus T_2}$

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:

$\ds S \setminus \bigcup_{i \mathop = 1}^k T_i = \bigcap_{i \mathop = 1}^k \paren {S \setminus T_i}$

Induction Step

Now we need to show:

$\ds S \setminus \bigcup_{i \mathop = 1}^{k + 1} T_i = \bigcap_{i \mathop = 1}^{k + 1} \paren {S \setminus T_i}$

This is our induction step:

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

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


$\ds S \setminus \bigcup_{i \mathop = 1}^n T_i = \bigcap_{i \mathop = 1}^n \paren {S \setminus T_i}$

that is:

$\ds S \setminus \bigcup_{i \mathop \in I} T_i = \bigcap_{i \mathop \in I} \paren {S \setminus T_i}$


Source of Name

This entry was named for Augustus De Morgan.