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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 $.

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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: $$ \{x\in A\mid A\text{ is a set with at most $2$ elements}\} $$

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  • $\begingroup$ 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 $\cup_{d\in D} P_d\times P_d$. $\endgroup$ – tomasz Feb 22 at 23:12

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