We now examine some of the basic operations that can be done to sets. The good thing about some of these operations is that they don’t require the use of any axioms, and can be formalized in logic alone. First we look at intersections.
Intersections
The intersection of two sets A and B is the set of all elements that belong to both A and B. For example,
\{1,2\} \cup \{2,3\} = \{2\}.
We start with our definition of a intersection.
For any two sets A and B, the intersection C is the set whose members are the members in both A and B.
A \cap B = \{x \mid x \in A \land x \in B\}
But of course before we can use our definition, we must prove the intersection exists and is unique.
For any two sets A and B, their intersection exists and is unique. Formally
(\forall A)(\forall B)(\exists!C)(\forall x)(x \in C \iff x \in A \land x \in B)
We are immediately given that the intersection exists by the axiom of specification. Given two sets A and B
(\exists C)(\forall x)(x \in C \iff x \in A \land x \in B).
In order to show it is unique, we use the axiom of extension. Let C^\prime be a second set such that
x \in C^\prime \iff x \in A \land x \in B.
Then for every x
x \in C^\prime \iff x \in C.
Thus by the axiom of extensionality
C^\prime = C.
Properties
We start of course by turning our definition into a theorem so we can use it in future theorems.
For all sets A and B, x \in A \cap B if and only if x \in A and x \in B.
We first convert our set builder notation into logic.
A \cap B = C \iff (\forall c)(c \in C \iff c \in A \land c \in B)
Then x \in A \cap B implies x \in C which implies x \in A and x \in B.
Now we assert the commutativity and associativity of intersection.
For all sets A and B, A \cap B = B \cap A
For all x \in A \cap B, x \in A and x \in B by Theorem 12. Since x \in B and x \in A. Also by Theorem 12 we have x \in B \cap A. So A \cap B \subseteq B \cap A.
Applying the same logic on B \cap A we have B \cap A \subseteq A \cap B. Thus A \cap B = B \cap A.
For all sets A, B and C, (A \cap B) \cap C = A \cap ( B \cap A).