Set Difference Union Second Set is Union

From ProofWiki
Jump to navigation Jump to search

Theorem

The union of a set difference with the second set is the union of the two sets.

That is, let $S, T$ be sets.

Then:

$\left({S \setminus T}\right) \cup T = S \cup T$


Proof

Consider $S, T \subseteq \mathbb U$, where $\mathbb U$ is considered as the universe.


\(\displaystyle \left({S \setminus T}\right) \cup T\) \(=\) \(\displaystyle \left({S \cap \complement \left({T}\right)}\right) \cup T\) Set Difference as Intersection with Complement
\(\displaystyle \) \(=\) \(\displaystyle \left({S \cup T}\right) \cap \left({\complement \left({T}\right) \cup T}\right)\) Union Distributes over Intersection
\(\displaystyle \) \(=\) \(\displaystyle \left({S \cup T}\right) \cap \mathbb U\) Union with Complement
\(\displaystyle \) \(=\) \(\displaystyle S \cup T\) Intersection with Universe

$\blacksquare$

Law of the Excluded Middle

This theorem depends on the Law of the Excluded Middle, by way of Union with Complement.

This is one of the axioms of logic that was determined by Aristotle, and forms part of the backbone of classical (Aristotelian) logic.

However, the intuitionist school rejects the Law of the Excluded Middle as a valid logical axiom. This in turn invalidates this theorem from an intuitionistic perspective.


Sources

This problem is worded: Show that in general $A - B \cup B \ne A$.