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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?

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  • $\begingroup$ 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. $\endgroup$
    – Mauro ALLEGRANZA
    Commented May 9, 2020 at 10:23

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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 $8\notin\{1,2\}$.

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    $\begingroup$ Well, $\{a\mid P(a)\}$ is the class of all objects ... $\endgroup$
    – Asaf Karagila ♦
    Commented May 7, 2020 at 18:03
  • $\begingroup$ @Asaf: Go to the head of the class! :-) $\endgroup$
    – Brian M. Scott
    Commented May 7, 2020 at 18:04
  • $\begingroup$ 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? $\endgroup$
    – koralgooll
    Commented May 7, 2020 at 20:46
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    $\begingroup$ @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. $\endgroup$
    – Brian M. Scott
    Commented Nov 3, 2020 at 18:18
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    $\begingroup$ @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. $\endgroup$
    – Brian M. Scott
    Commented Nov 3, 2020 at 19:22

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