Union with Relative Complement

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Theorem

The union of a set $T$ and its relative complement in $S$ is the set $S$:

$\relcomp S T \cup T = S$


Proof

Step 1

By the definition of relative complement, we have that:

$\relcomp S T \subseteq S$

and:

$T \subseteq S$

Hence by Union is Smallest Superset:

$\relcomp S T \cup T\subseteq S$

$\Box$


Step 2

Let $x \in S$.

By Law of Excluded Middle, one of the following two applies:

$(1): \quad x \in T$
$(2): \quad x \notin T$


If $(2)$, then by definition of relative complement:

$x \in S \setminus T = \relcomp S T$

So:

$x \in T \lor x \in \relcomp S T$

By definition of set union:

$x \in \relcomp S T \cup T$

Thus:

$x \in S \implies x \in \relcomp S T \cup T$

By definition of subset it follows that:

$S \subseteq \relcomp S T \cup T$

$\Box$


Step 3

From:

$\relcomp S T \cup T\subseteq S$

and:

$S \subseteq \relcomp S T \cup T$

it follows from definition of set equality that:

$S = \relcomp S T \cup T$

$\blacksquare$


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.


Also see


Sources

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