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

From ProofWiki
Jump to navigation Jump to search

Theorem

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

where:

$\ds \bigcap \mathbb T := \set {x: \forall T' \in \mathbb T: x \in T'}$

that is, the intersection of $\mathbb T$


Proof

Suppose:

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

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

Then:

\(\ds x\) \(\in\) \(\ds S \setminus \bigcap \mathbb T\)
\(\ds \leadstoandfrom \ \ \) \(\ds x\) \(\notin\) \(\ds \bigcap \mathbb T\) Definition of Set Difference
\(\ds \leadstoandfrom \ \ \) \(\ds \neg \leftparen {\forall T' \in \mathbb T}: \, \) \(\ds x\) \(\in\) \(\ds \rightparen {T'}\) Definition of Intersection of Set of Sets
\(\ds \leadstoandfrom \ \ \) \(\ds \exists T' \in \mathbb T: \, \) \(\ds x\) \(\notin\) \(\ds T'\) Denial of Universality
\(\ds \leadstoandfrom \ \ \) \(\ds \exists T' \in \mathbb T: \, \) \(\ds x\) \(\in\) \(\ds S \setminus T'\) Definition of Set Difference: note $x \in S$ from above
\(\ds \leadstoandfrom \ \ \) \(\ds x\) \(\in\) \(\ds \bigcup_{T' \mathop \in \mathbb T} \paren {S \setminus T'}\) Definition of Union of Set of Sets


Therefore:

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

$\blacksquare$


Source of Name

This entry was named for Augustus De Morgan.


Sources