The Intersection of Sets is a Set

2018 January 13


Modern mathematics is almost universally* based on set theory, so I decided to read Pinter's Set Theory.

Naive set theory says that a set is any collection of objects. For example, the set of natural numbers, , is the set {1, 2, 3, …}. Naive set theory runs into several problems, so most mathematicians use Zermelo-Fraenkel (ZF) set theory, so this post also uses it. The difference is not important for this post.

The intersection A ∩ B of 2 sets, A and B, is the set of objects which are in both A and B. For example,
{1, 2, 3}∩{2, 3, 4}={2, 3}

Note that the associative property holds for . Also,
$$\bigcap_{i \in I}^{} A_i = A_1 \cap A_2 \cap \ldots \cap A_n$$

Here is a proof of an exercise in Set Theory.


Conjecture

Let {Ai}i ∈ I be an indexed family of sets. Prove that
$$\bigcap_{i \in I}^{} A_i$$
is a set.

Proof

By definition, the intersection of 2 sets is a set. The associative property holds for the intersection, so
$$\bigcap_{i \in I}^{} A_i = A_1 \cap (A_2 \cap \ldots \cap A_n)$$
We already know that A1 is a set, so $\bigcap_{i \in I}^{} A_i$ is a set if (A2 ∩ … ∩ An) is a set. We can repeat this step, writing that
A2 ∩ … ∩ An = A2 ∩ (A3 ∩ …An)
Again, we see that A2 is a set, so we now need to prove A3 ∩ …An is a set. We repeat this step until we get
An − 1 ∩ (An)
which is a set, so now we go backwards, seeing that
An − 2 ∩ (An − 1 ∩ An)
is a set, and continuing backwards until we prove that
A1 ∩ (A2 ∩ … ∩ An)
is a set, which equals
$$\bigcap_{i \in I}^{} A_i$$
Thus, $\bigcap_{i \in I}^{} A_i$ is a set. QED


Notes