Union with Relative Complement

Theorem

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

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

Proof

Step 1

By the definition of relative complement, we have that $\complement_S \left({T}\right)\subseteq S$ and $T \subseteq S$.

Hence by Union Smallest, $\complement_S \left({T}\right) \cup T\subseteq S$.

$\Box$

Step 2

Let $x \in S$.

By the Law of the 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, $x \in S \setminus T = \complement_S \left({T}\right)$.

So $x \in T \lor x \in \complement_S \left({T}\right)$.

By definition of set union, $x \in \complement_S \left({T}\right) \cup T$.

Thus $x \in S \implies x \in \complement_S \left({T}\right) \cup T$.

By definition of subset it follows that $S \subseteq \complement_S \left({T}\right) \cup T$.

$\Box$

Step 3

From:

$\complement_S \left({T}\right) \cup T\subseteq S$

and:

$S\subseteq \complement_S \left({T}\right) \cup T$

it follows from Equality of Sets that $S = \complement_S \left({T}\right) \cup T$.

$\blacksquare$

Note

This proof depends directly on the Law of the Excluded Middle.

This rule is denied by the intuitionist school.

Sources

• Paul R. Halmos: Naive Set Theory (1960)... (previous)... (next): $\S 5$: Complements and Powers
• Seth Warner: Modern Algebra (1965)... (previous)... (next): $\S 3$: Theorem $3.2$
• T.S. Blyth: Set Theory and Abstract Algebra (1975): $\S 1$
• Keith Devlin: The Joy of Sets: Fundamentals of Contemporary Set Theory (1993): $\S 1.2$: Exercise $1.2.2 \ \text{(iii)}$