# Specific question about example of Family of sets & Set builder notation

I encountered the following example whilst studying Equivalence Classes from How to prove it, Velleman. :

Let $$B= \{(p, q) ∈ P × P | \text{the person p has the same birthday as the person q } \}$$

Let D be the set of all possible birthdays. In other words,$$D = \{Jan. 1, Jan. 2, Jan. 3, . . . ,Dec. 30, Dec. 31\}.$$

Now for each $$d ∈ D$$, let $$P_d = \{p ∈ P | \text{the person p was born on the day d } \}$$ .

Then the family $$F = \{P_d | d ∈ D\}$$ is an indexed family of subsets of P.

We can then redefine B as:

$$B = \{(p, q) ∈ P × P | ∃d ∈ D(p ∈ P_d \land q ∈ P_d )\}$$

$$= \{(p, q) ∈ P × P | ∃d ∈ D((p, q) ∈ P_d × P_d )\}$$

$$= ∪_{d∈D}(P_d × P_d ).$$

My question: Could I write $$\{(p, q) ∈ P × P | ∃d ∈ D((p, q) ∈ P_d × P_d )\}$$ as $$\{(p, q) ∈ P_d × P_d | d ∈ D\}$$?

I think yes ( but I want to be sure ). My reasoning is since inside the set builder notation we have $$(p,q) ∈ P × P \land (p,q) ∈ P_d × P_d$$ then $$(P × P) \land (P_d × P_d) = (P_d × P_d)$$ Since $$P_d × P_d \subseteq P × P$$.

This is not standard set builder notation and I would advise against using it. The general scheme is: $$B:=\{x\in A\mid\varphi(x)\}$$ Here the collection $$A$$ serves as an upper bound on where the $$x$$'s come from, so that you immediately know that $$B\subseteq A$$. This notation is a direct implementation of the axiom of restricted comprehension, which tells us that if $$A$$ is a set, then $$B$$ as defined above is a set as well.
Crucially, the set $$A$$ is fixed before defining $$B$$! In your notation $$\{(p, q) ∈ P_d × P_d | d ∈ D\}$$ however, the role of $$A$$ is taken by $$P_d \times P_d$$ which depends on the parameter $$d$$ which in turn varies over $$D$$. While it is clear what you want to define, your relaxed notation makes it very easy to write down set builder terms which denote proper classes, for example: $$\{x\in A\mid A\text{ is a set with at most 2 elements}\}$$
• I think it is worth noting that you can write the set in a much more concise way without using the set builder notation at all. It is simply $\cup_{d\in D} P_d\times P_d$. – tomasz Feb 22 at 23:12