Set Difference with Empty Set is Self

Theorem

The set difference between a set and the empty set is the set itself:

$S \setminus \O = S$

Proof

From Set Difference is Subset:

$S \setminus \O \subseteq S$

From the definition of the empty set:

$\forall x \in S: x \notin \O$

Let $x \in S$.

Thus:

\(\ds x\)

\(\in\)

\(\ds S\)

\(\ds \leadsto \ \ \)

\(\ds x\)

\(\in\)

\(\ds S \land x \notin \O\)

Rule of Conjunction

\(\ds \leadsto \ \ \)

\(\ds x\)

\(\in\)

\(\ds S \setminus \O\)

Definition of Set Difference

\(\ds \leadsto \ \ \)

\(\ds S\)

\(\subseteq\)

\(\ds S \setminus \O\)

Definition of Subset

Thus we have:

$S \setminus \O \subseteq S$

and:

$S \subseteq S \setminus \O$

So by definition of set equality:

$S \setminus \O = S$

$\blacksquare$

Also see

• Set Difference with Self is Empty Set

Sources

• 1964: W.E. Deskins: Abstract Algebra ... (previous) ... (next): Exercise $1.1: \ 8 \ \text{(b)}$
• 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $1$: The Notation and Terminology of Set Theory: $\S 8 \ \text{(d)}$
• 1993: Keith Devlin: The Joy of Sets: Fundamentals of Contemporary Set Theory (2nd ed.) ... (previous) ... (next): $\S 1$: Naive Set Theory: $\S 1.2$: Operations on Sets: Exercise $1.2.5 \ \text{(i)}$
• 2012: M. Ben-Ari: Mathematical Logic for Computer Science (3rd ed.) ... (previous) ... (next): Appendix $\text{A}.2$: Theorem $\text{A}.11$