# Set builder notation used in intersection of x?

I would like to ask about notation, because I think that I am missing something.

It is from book; "The Joy of Sets" - Keith Devlin. There is presented an alternative notation; $$x = \{a | P(a) \}$$ which means that $$x$$ is a set of all those $$a$$ for which $$P(a)$$ holds. Does it mean that $$P(a)$$ has to evaluate to true?

Then we have definition of intersection of $$x$$ which looks like that $$\bigcap x = \{ a | \forall y (y \in x \rightarrow a \in y ) \}$$. Let's assume that $$x = \{ \{ 1, 2 \}, \{ 1, 3 \} \}$$ then $$\bigcap x$$ should be equal to $$\{ 1 \}$$ but I can also said that it is equal $$\{ 8 \}$$, because $$y = \{ 8 \}$$ is not an element of $$x$$ so the $$P(a)$$ is evaluated to true, so $$\bigcap x = \{ 8 \}$$ which is wrong. Can you explain me what I am wrongly mixing? I know that normally I would assume that $$y \in x$$ is true, but in this case the $$P(a)$$ should be true, shouldn't it?

• The "rule" with the set-builder notation is: $z \in \{ a \mid P(a) \} \text { iff } P(z)$. Thus, $8 \in \bigcap x \text { iff } \forall y (y \in x \to 8 \in y)$. The $y$ in $x$ are $\{ 1,2 \}$ and $\{ 1,3 \}$ and $8$ does not belongs to neither of them. Commented May 9, 2020 at 10:23

## 1 Answer

Yes, when we write $$x=\{a\mid P(a)\}$$, we mean that $$x$$ is the set of all objects $$a$$ such that $$P(a)$$ is a true statement.

You have $$\bigcap x$$ defined to be $$\{a\mid\forall y\,(y\in x\to a\in y)\}$$: $$\bigcap x$$ is the set of all objects $$a$$ for which it is true that $$a\in y$$ whenever $$y\in x$$. In other words, the definition say precisely that $$\bigcap x$$ is the set of all objects that are elements of every element of $$x$$. When $$x=\{\{1,2\},\{1,3\}\}\;,$$ this means that $$\bigcap x=\{1\}$$: it is true that $$\forall y\,(y\in x\to 1\in y)$$, and there is no other object $$a$$ of which we can truly say that $$\forall y\,(y\in x\to a\in y)$$. In particular, $$\forall y\,(y\in x\to 8\in y)$$ is not true, because $$\{1,2\}\in x$$, but $$8\notin\{1,2\}$$.

• Well, $\{a\mid P(a)\}$ is the class of all objects ... Commented May 7, 2020 at 18:03
• @Asaf: Go to the head of the class! :-) Commented May 7, 2020 at 18:04
• Thank You @BrianM.Scott but in your solution, you assume true in the first step $\{ 1, 2 \} \in x$. I have a little exted my knowledge and I start think it is because of $\forall y$, which sets the domain, isn't it? If it is true why the elements of domain are $\{ 1, 2 \}$ and $\{ 1, 3 \}$? What should I read to understand it well? Something about first-order logic? Commented May 7, 2020 at 20:46
• @TaylorRendon: I’d take the whole thing to be the definition: $E$ limits the domain from which we’re selecting the members of $A$, and $\varphi$ gives their defining characteristic. It’s true that $E$ is sometimes a bit artificial, present largely to satisfy the requirements of the comprehension axiom schema, but sometimes it’s a real limitation: $\{x\in\Bbb N:3\le x\le 9\}$ is very different from $\{x\in\Bbb R:3\le x\le 9\}$, for instance. Commented Nov 3, 2020 at 18:18
• @TaylorRendon: Yes. Think about how you’d say the whole thing if you weren’t specifically trying to follow the syntax of the symbols: the set of real numbers between $3$ and $9$ inclusive, for instance. Commented Nov 3, 2020 at 19:22