Proving the uniqueness of the empty set

While proving some assertion about sets, the need for showing that a set is equals to the empty set may arise. In that situation, a lot of confusion appear if the uniqueness of the empty set is unknown, because showing that the empty set, which we know has no element, is equal to another different, but also with no element, makes no sense, since the empty set is just one, the $\emptyset$.

So, we can claim that there's only one set with no elements.

Proof: Suppose $A, B$ are sets with no elements. It's known that $A \subset B$, and $B \subset A$. Then, by definition of equality, $A = B$.

In this way, in order to show that some set is equals to the empty set, it's necessary only to show that the set has no elements, and, due to the uniqueness, it will be undoubtedly the empty set $\emptyset$.

Proving properties of set's operations

Defining operations on sets

In order to prove some properties of set's operations, it's worth precisely defining those operations:

• Union: $A \cup B = \{x \mid x \in A \lor x \in B\}$
• Intersection: $A \cap B = \{x \mid x \in A \land x \in B\}$
• Difference: $A - B = \{x \mid x \in A \land x \not \in B\}$
• Complement: $A^\complement = \{x \mid x \not \in A\}$

Summary

1. Comutative Laws
2. Associtative Laws
3. Distributive Laws
4. Identity Laws
5. Complement Laws
6. Double Complement Laws
7. Idempotent Laws
8. Universal Bound Laws
9. De Morgan's Laws
10. Absorption Laws
11. Complements of U and $\emptyset$
12. Set Difference Law

Proving properties

1 - Commutative Laws

a) $A \cup B = B \cup A$

Let $A, B$ be arbitrary sets.
Suppose $x \in A \cup B$.
Then, by definition, $x \in A \lor x \in B$.
Case 1: if $x \in A$, $x \in B \lor x \in A$.
Case 2: if $x \in B$, $x \in B \lor x \in A$.
So, $\forall x \in A \cup B$, $x \in B \lor x \in A$.
By definition, $x \in B \cup A$.
$\forall A,B$, if $x \in A \cup B$, then $x \in B \cup A$.

Let $A, B$ be arbitrary sets.
Suppose $x \in B \cup A$.
Then, by definition, $x \in B \lor x \in A$.
Case 1: if $x \in B$, $x \in B \lor x \in A$.
Case 2: if $x \in A$, $x \in B \lor x \in A$.
So, $\forall x \in B \cup A$, $x \in A \lor x \in B$.
By definition, $x \in A \cup B$.
$\forall A,B$, if $x \in B \cup A$, then $x \in A \cup B$.

$\forall x \in A \cup B$, $ x \in B \cup A$.
$\forall x \in B \cup A$, $x \in A \cup B$.
By definition, $A \cup B = B \cup A$.

b) $A \cap B = B \cap A$

Let $A, B$ be arbitrary sets.
Suppose $x \in A \cap B$.
Then, by definition, $x \in A \land x \in B$ $x \in B$
$x \in A$
Then, $x \in B \land x \in A$
So, $\forall x \in A \cap B, x \in B \land x \in A$
By definition, $x \in B \cap A$
$\forall A,B$, if $x \in A \cap B$, then $x \in B \cap A$.

Let $A, B$ be arbitrary sets.
Suppose $x \in B \cap A$.
Then, by definition, $x \in B \land x \in A$ $x \in A$
$x \in B$
Then, $x \in A \land x \in B$
So, $\forall x \in B \cap A, x \in A \land x \in B$
By definition, $x \in A \cap B$
$\forall A,B$, if $x \in B \cap A$, then $x \in A \cap B$.

$\forall A,B$, if $x \in A \cap B$, $x \in B \cap A$.
$\forall A,B$, if $x \in B \cap A$, $x \in A \cap B$.
By definition, $A \cap B = B \cap A$.

2 - Associative Laws

a) $(A \cup B) \cup C = A \cup (B \cup C)$

Let $A, B, C$ be arbitrary sets.
Suppose $x \in (A \cup B) \cup C$
Then $x \in (A \cup B) \lor x \in C$
Then $(x \in A \lor x \in B) \lor x \in C$.
Case 1: if $x \in A$, $x \in A \lor (x \in B \lor x \in C)$.
Then $x \in A \lor x \in A \cup B$, then $x \in A \cup (B \cup C)$.
Case 2: if $x \in B$, $x \in B \lor x \in C$. If $x \in B \lor x \in C, (x \in B \lor x \in C) \lor x \in A.$
Then $x \in B \cup C \lor x \in A$, then $x \in (B \cup C) \cup A$.
By commutative law, $x \in A \cup (B \cup C)$.
Case 3: if $x \in C$, $x \in B \lor x \in C$. If $x \in B \lor x \in C, (x \in B \lor x \in C) \lor x \in A.$
Then $x \in B \cup C \lor x \in A$, then $x \in (B \cup C) \cup A$.
By commutative law, $x \in A \cup (B \cup C)$.
$\forall A,B,C$, if $x \in (A \cup B) \cup C$, then $x \in A \cup (B \cup C)$.

Using the same reasoning, supposing $x \in A \cup (B \cup C)$ we conclude that
$\forall A,B,C$, if $x \in A \cup (B \cup C)$, then $x \in (A \cup B) \cup C$.

$\forall A,B,C$, if $x \in (A \cup B) \cup C$, then $x \in A \cup (B \cup C)$.
$\forall A,B,C$, if $x \in A \cup (B \cup C)$, then $x \in (A \cup B) \cup C$.
By definition, $(A \cup B) \cup C = A \cup (B \cup C)$.

b) $(A \cap B) \cap C = A \cap (B \cap C)$

Let $A, B, C$ be arbitrary sets.
Suppose $x \in (A \cap B) \cap C$.
Then $x \in (A \cap B) \land x \in C$.
Then $(x \in A \land x \in B) \land x \in C$.
$x \in B$.
$x \in C$.
$x \in B \land x \in C$, then $x \in B \cap C$.
$x \in A$, then $x \in A \land x \in (B \cap C)$.
By definition, $x \in A \cap (B \cap C)$.
So, $\forall A, B, C$, if $x \in (A \cap B) \cap C$, then $x \in A \cap (B \cap C)$.

Using the same reasoning, supposing $x \in A \cap (B \cap C)$ we conclude that
$\forall A, B, C$, if $x \in A \cap (B \cap C)$, then $x \in (A \cap B) \cap C$.
By definition, $(A \cap B) \cap C = A \cap (B \cap C)$.

3 - Distributive Laws

a)$A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$

Let $A, B, C$ be arbitrary sets.
Suppose $x \in A \cup (B \cap C)$.
Then $x \in A \lor x \in (B \cap C).$
Case 1: if $x \in A$, $x \in A \lor x \in B$, then $x \in A \cup B$.
Also, if $x \in A$, $x \in A \lor x \in C$, then $x \in A \cup C$.
Then, if $x \in A$, $x \in A \cup B \land x \in A \cup C$.
By definition, $x \in (A \cup B) \cap (A \cup C).$
$\forall A,B,C$, if $x \in A \cup (B \cap C)$, then $x \in (A \cup B) \cap (A \cup C)$.

Let $A, B, C$ be arbitrary sets.
Suppose $x \in (A \cup B) \cap (A \cup C).$
Then $x \in (A \cup B) \land x \in (A \cup C).$
Then $(x \in A \lor x \in B) \land (x \in A \lor x \in C).$
Case 1: If $x \in A$, $x \in A \lor x \in (B \cap C)$, then $x \in A \cup (B \cap C)$.
Case 2: If $x \not \in A$, $x \in B \land x \in C$, then $x \in (B \cap C)$.
If $x \in (B \cap C)$, $x \in A \lor x \in (B \cap C)$.
By definition, $x \in A \cup (B \cap C)$.
$\forall A,B,C$, if $x \in (A \cup B) \cap (A \cup C)$, then $x \in A \cup (B \cap C)$.

$\forall A,B,C$, if $x \in A \cup (B \cap C)$, then $x \in (A \cup B) \cap (A \cup C)$.
$\forall A,B,C$, if $x \in (A \cup B) \cap (A \cup C)$, then $x \in A \cup (B \cap C)$.
By definition, $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$.

b)$A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$

Let $A, B, C$ be arbitrary sets.
Suppose $x \in A \cap (B \cup C)$.
Then $x \in A \land x \in (B \cup C).$
$x \in A$.
$x \in (B \cup C).$
Then, $x \in B \lor x \in C$.
Case 1: If $x \in B, x \in A \land x \in B$, then $x \in A \cap B$.
Case 2: If $x \in C, x \in A \land x \in C$, then $x \in A \cap C$.
Then, $x \in A \cap B \lor x \in A \cap C$.
By definition, $x \in (A \cap B) \cup (A \cap C).$
$\forall A,B,C$, if $x \in A \cap (B \cup C)$, then $x \in (A \cap B) \cup (A \cap C)$.

Let $A, B, C$ be arbitrary sets.
Suppose $x \in (A \cap B) \cup (A \cap C)$.
Then $x \in (A \cap B) \lor x \in (A \cap C)$.
Case 1: if $x \in (A \cap B), x \in A \land x \in B$.
If $x \in B, x \in B \lor x \in C$, then $x \in B \cup C$.
Given that $x \in A \land x \in B \cup C, x \in A \cap (B \cup C)$.
Case 2: if $x \in (A \cap C), x \in A \land x \in C$.
If $x \in C, x \in B \lor x \in C$, then $x \in B \cup C$.
Given that $x \in A \land x \in B \cup C, x \in A \cap (B \cup C)$.
$\forall A,B,C$, if $x \in (A \cap B) \cup (A \cap C)$, then $x \in A \cap (B \cup C)$.

$\forall A,B,C$, if $x \in A \cap (B \cup C)$, then $x \in (A \cap B) \cup (A \cap C)$.
$\forall A,B,C$, if $x \in (A \cap B) \cup (A \cap C)$, then $x \in A \cap (B \cup C)$.
By definition, $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$.

4 - Identity Laws

a) $A \cup \emptyset = A$

Let $A$ be an arbitrary set.
Suppose $x \in A \cup \emptyset$.
Then $x \in A \lor x \in \emptyset$.
Case 1: if $x \in A, x \in A$.
Case 2: if $x \not \in A, x \in \emptyset$. Absurd, then $x \in A$.
$\forall A$, if $x \in A \cup \emptyset$, then $x \in A$.

Let $A$ be an arbitrary set.
Suppose $x \in A$.
If $x \in A$, $x \in A \lor x \in \emptyset$, then $x \in A \cup \emptyset$.
$\forall A$, if $x \in A, x \in A \cup \emptyset$.

$\forall A$, if $x \in A \cup \emptyset$, then $x \in A$.
$\forall A$, if $x \in A, x \in A \cup \emptyset$.
By definition, $A \cup \emptyset = A$.

b) $A \cap U = A$

Let $A$ be an arbitrary set and $A \subset U$.
Suppose $x \in A \cap U$.
Then $x \in A \land x \in U$.
$x \in A$.
$\forall A$, if $x \in A \cap U$, then $x \in A$.

Let $A$ be an arbitraty set and $A \subset U$.
Then $\forall x \in A, x \in U$.
Suppose $x \in A$, then $x \in U$, then $x \in A \cap U$.
$\forall A$, if $x \in A$, then $x \in A \cap U$.

$\forall A$, if $x \in A \cap U$, then $x \in A$.
$\forall A$, if $x \in A$, then $x \in A \cap U$.
By definition, $A \cap U = A$.

5 - Complement Laws

a) $A \cup A^\complement = U$

Let $A$ be an arbitrary set and $A \subset U$.
Suppose $x \in A \cup A^\complement$.
Then $x \in A \lor x \in A^\complement$.
$x \in U$.
Case 1: if $x \in A$, then $x \in U$.
Case 2: if $x \in A^\complement$, $x \not \in A$, $x \in U$.
$\forall A$, if $x \in A \cup A^\complement$, then $x \in U$.

Let $A$ be an arbitrary set and $A \subset U$.
Suppose $x \in U$.
$x \in A \lor x \not \in A$.
So, $x \in A \cup A^\complement$.
$\forall A$, if $x \in U$, then $x \in A \cup A^\complement$.

$\forall A$, if $x \in A \cup A^\complement$, then $x \in U$.
$\forall A$, if $x \in U$, then $x \in A \cup A^\complement$.
By definition, $A \cup A^\complement = U$.

b) $A \cap A^\complement = \emptyset$

Let $A$ be an arbitraty set.
Suppose $x \in A \cap A^\complement$.
Then $x \in A \land x \not \in A$. Absurd.
$\forall A, \not \exists x \mid x \in A \cap A^\complement$.
By the the definition of an empty set and the uniqueness of the empty set, $A \cap A^\complement = \emptyset$.

6 - Double Complement Law

a)$(A^\complement)^\complement = A$

Let $A$ be an arbitrary set.
Suppose $x \in (A^\complement)^\complement$. Then $x \not \in A^\complement$. Then $x \in A$. $\forall A$, if $x \in (A^\complement)^\complement$, then $x \in A$.
Let $A$ be an arbitrary set.
Suppose $x \in A$.
Then $x \not \in A^\complement$.
By definition, $x \in (A^\complement)^\complement$.
$\forall A$, if $x \in A$, then $x \in (A^\complement)^\complement$.
$\forall A$, if $x \in (A^\complement)^\complement$, then $x \in A$. $\forall A$, if $x \in A$, then $x \in (A^\complement)^\complement$. By definition, $(A^\complement)^\complement = A$.

7 - Idempotent Laws

a)$A \cup A = A$

Let $A$ be an arbitrary set.
Suppose $x \in A \cup A$.
Then $x \in A \lor x \in A$.
$x \in A$.
$\forall A$, if $x \in A \cup A$, then $x \in A$.

Let $A$ be an arbitrary set.
Suppose $x \in A$.
Then $x \in A \lor x \in A$.
By definition, $x \in A \cup A$.
$\forall A$, if $x \in A$, then $x \in A \cup A$.

$\forall A$, if $x \in A \cup A$, then $x \in A$.
$\forall A$, if $x \in A$, then $x \in A \cup A$.
By definition, $A \cup A = A$.

b)$A \cap A = A$

Let $A$ be an arbitrary set.
Suppose $x \in A \cap A$.
Then $x \in A \land x \in A$.
$x \in A$.
$\forall A$, if $x \in A \cap A$, then $x \in A$.

Let $A$ be an arbitrary set.
Suppose $x \in A$.
Then $x \in A \land x \in A$.
By definition, $x \in A \cap A$.
$\forall A$, if $x \in A$, then $x \in A \cap A$.

$\forall A$, if $x \in A \cap A$, then $x \in A$.
$\forall A$, if $x \in A$, then $x \in A \cap A$.
By definition, $A \cap A = A$.

8 - Universal Bound Laws

a)$A \cup U = U$

Let $A$ be an arbitrary set and $A \subset U$.
By definition of subset, $\forall x \in A, x \in U$.
Suppose $x \in A \cup U$.
Then $x \in A \lor x \in U$.
Case 1: if $x \in A, x \in U$.
Case 2: if $x \in U, x \in U$.
$\forall x \in A \cup U, x \in U$.

Let $A$ be an arbitrary set and $A \subset U$.
By definition of subset, $\forall x \in A, x \in U$.
Suppose $x \in U$.
Then $x \in A \lor x \in U$, then $x \in A \cup U$.
$\forall x \in A, x \in A \cup U$.

$\forall x \in A \cup U, x \in U$.
$\forall x \in A, x \in A \cup U$.
By definition, $A \cup U = U$.

b)$A \cap \emptyset = \emptyset$

Let $A$ be an arbitrary set.
Suppose $x \in A \cap \emptyset$.
Then $x \in A \land x \in \emptyset$.
$x \in \emptyset$. Absurd.
Then $\not \exists x \mid x \in A \cap \emptyset$.

By the the definition of an empty set and the uniqueness of the empty set, $A \cap \emptyset = \emptyset$.

9 - De Morgan's Laws

a)$(A \cup B)^\complement = A^\complement \cap B^\complement$

b)$(A \cap B)^\complement = A^\complement \cup B^\complement$

10 - Absorption Laws

a)$A \cup (A \cap B) = A$

Let $A, B$ be arbitrary sets.
Suppose $x \in A \cup (A \cap B)$.
Then $x \in A \lor x \in (A \cap B)$.
Case 1: if $x \in A$, then $x \in A$.
Case 2: if $x \in A \cap B$, then $x \in A \land x \in B$, then $x \in A$.
$\forall A$, if $x \in A \cup (A \cap B)$, then $x \in A$.

Let $A, B$ be arbitrary sets.
Suppose $x \in A$.
Then $x \in A \lor x \in A \cup (A \cap B)$.
$\forall A$, if $x \in A$, then $x \in A \cup (A \cap B)$.

$\forall A$, if $x \in A \cup (A \cap B)$, then $x \in A$.
$\forall A$, if $x \in A$, then $x \in A \cup (A \cap B)$.
By definition, $A \cup (A \cap B) = A$.

b)$A \cap (A \cup B) = A$

Let $A, B$ be arbitrary sets.
Suppose $x \in A \cap (A \cup B)$.
Then $x \in A \land x \in (A \cup B)$.
$x \in A$.
$\forall A$, if $x \in A \cap (A \cup B)$, then $x \in A$.

Let $A, B$ be arbitrary sets.
Suppose $x \in A$.
Then $x \in A \lor x \in B$, then $x \in (A \cup B)$.
Then $x \in A \land x \in (A \cup B)$, then $x \in A \cap (A \cup B)$.
$\forall A$, if $x \in A$, then $x \in A \cap (A \cup B)$.

$\forall A$, if $x \in A \cap (A \cup B)$, then $x \in A$.
$\forall A$, if $x \in A$, then $x \in A \cap (A \cup B)$.
By definition, $A \cap (A \cup B) = A$.

11 - Complements of U and $\emptyset$

a)$U^\complement = \emptyset$

Let $A$ be an arbitrary set and $A \subset U$.
So, $\forall x \in A, x \in U$.
Suppose $x \in U^\complement$.
Then, $x \not \in U$. Absurd.
Then, $\not \exists x \mid x \in U^\complement$.

By the the definition of an empty set and the uniqueness of the empty set, $U^\complement = \emptyset$.

b)$\emptyset^\complement = U$

Suppose an arbitrary $x \in \emptyset^\complement$.
Then $x \not \in \emptyset$.
Then $x \in U$.
$\forall x \in \emptyset^\complement$, $x \in U$.

Suppose an arbitrary $x \in U$.
Then $x \not \in \emptyset$.
Then $x \in \emptyset^\complement$.
$\forall x \in U$, then $x \in \emptyset^\complement$.

$\forall x \in \emptyset^\complement$, $x \in U$.
$\forall x \in U$, then $x \in \emptyset^\complement$.
By definition, $\emptyset^\complement = U$.

12 - Set Difference Law

a)$A - B = A \cap B^\complement$

Let $A, B$ be arbitrary sets.
Suppose $x \in A - B$.
Then $x \in A \land x \not \in B$.
$x \in A$.
By definition, if $x \not \in B$, then $x \in B^\complement$.
Then, $x \in A \land x \in B^\complement$, then $x \in A \cap B^\complement$.
$\forall A, B$, if $x \in A - B$, then $x \in A \cap B^\complement$.

Let $A, B$ be arbitrary sets.
Suppose $x \in A \cap B^\complement$.
Then, $x \in A \land x \in B^\complement$.
If $x \in B^\complement$, then $x \not \in B$.
$x \in A$.
$If x \in A \land x \not \in B$, by definition, $x \in A - B$.
$\forall A, B$, if $x \in A \cap B^\complement$, $x \in A - B$.

$\forall A, B$, if $x \in A - B$, then $x \in A \cap B^\complement$.
$\forall A, B$, if $x \in A \cap B^\complement$, $x \in A - B$.
By definition, $A - B = A \cap B^\complement$.