Relative Complement of Relative Complement

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Theorem

The relative complement of the relative complement of a set is itself:

$\relcomp S {\relcomp S T} = T$


Thus, considered as a mapping on the power set of $S$:

$\complement_S: \powerset S \to \powerset S: \relcomp S T = S \setminus T$

$\complement_S$ is an involution.


Proof 1

By the definition of relative complement:

$\relcomp S {\relcomp S T} = S \setminus \paren {S \setminus T}$

Let $t \in T$.

Then by the definition of set difference:

$t \notin S \setminus T$

Since $T \in T$ and $T \subseteq S$, by the definition of subset:

$t \in S$

Thus:

$t \in \paren {S \setminus \paren {S \setminus T} }$


Suppose instead that:

$t \in \paren {S \setminus \paren {S \setminus T} }$

Then:

$t \in S$

and:

$\neg \paren {t \in \paren {S \setminus T} }$

Thus:

$\neg \paren {\paren {t \in S} \land \neg \paren {t \in T} }$

By Conditional is Equivalent to Negation of Conjunction with Negative:

$t \in S \implies t \in T$

By Modus Ponendo Ponens:

$t \in T$

$\blacksquare$


Proof 2

\(\ds \relcomp S {\relcomp S T}\) \(=\) \(\ds S \setminus \paren {S \setminus T}\) Definition of Relative Complement
\(\ds \) \(=\) \(\ds S \cap T\) Set Difference with Set Difference


The definition of the relative complement requires that:

$T \subseteq S$

But from Intersection with Subset is Subset‎:

$T \subseteq S \iff T \cap S = T$

Thus:

$\relcomp S {\relcomp S T} = T$

follows directly.

$\blacksquare$


Law of the Excluded Middle

This theorem depends on the Law of the Excluded Middle.

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

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