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In set theory

Axiom Schema of Separation: If $P$ is a property (with parameter$p$), then for any $X$ and $p$ there exists a set $Y = \lbrace u ∈ X : P(u, p) \rbrace$ that contains all those $u ∈ X$ that have property $P$.

But some of the books are written as follows:

Axiom Schema of Separation: If $P(u)$ is a property then for any $X$ and $P(u)$ there exists a set $Y = \lbrace u ∈ X : P(u) \rbrace$ that contains all those $u ∈ X$ that have property $P(u)$.

But I do't know that what's this $P (u)$ . what is its definition mathematically of $P(u)$ (and $P(u ,p_1,p_2,...,p_n)$?

Almot1960
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4 Answers4

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$P$ is a property: a statement that takes some parameter(s), and is either true or false, depending on those parameters, $P(u)$ is the result of evaluating $P$ with the parameter $u$ (and the same for any other collection of parameters that are valid inputs into $P$).

user3482749
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$p(u)$ can be anything, really, as long as it can be stated with some kind of logic formula.

For example, if $X$ is the set of natural numbers, then $p(u)$ could express the property that $u$ is an even number. But you could also have it express that $u$ is an odd number. Or that it is a prime number. Or that $u$ is a prime number that is the sum of three odd numbers. Or ... like I said, pretty much anything you want.

Note that if you use $u=u$ for the formula $p(u)$, then the subset $Y$ is just the same as $X$ itself, as $u=u$ is true for any $u$. And, if you use the formula $u \not = u$, you end up with the empty set.

Bram28
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  • I know that $f(x)$ is the value of $f$ (function) for the value $x$ of its variable. why symbol of property is to form $P(u)$ ? – Almot1960 Dec 15 '18 at 21:23
  • @Almot1960 That's just the notation that's used. There's definitely a reasonable complaint here, that the notation makes functions and properties/relations look overly similar, but oh well. – Noah Schweber Dec 15 '18 at 21:30
  • Thus $P(u)$ not related to the function .and only is a notation . it's true ? – Almot1960 Dec 15 '18 at 21:36
  • @Almot1960 $p(u)$ is indeed not a function .. it is a property (or predicate). – Bram28 Dec 15 '18 at 21:38
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The way I read this $P$ could be any property. "is an integer divisible by $7$" for example. Or "eats shoots and leaves".

All the schema is saying that given any set, there exists a subset of all elements with the property.

For example. If $P$ = "is divisible by $7$" and $Q$ = "eats shoots and leaves" and $A = $ set of all perfect squares. And $B =$ set of all black and white animals then there is $E = \{x\in A|P(x)\} =$ set of all perfect squares that are divisible by $7=\{49k^2| k\in \mathbb Z\}$.

There is a $F = \{x \in A|Q(x)\} =$ set of all perfect squares that eat shoots and leaves = $\emptyset$.

And there is $G = \{x\in B|P(x)\} = $ set of all black and white animals that are divisible by $7 = \emptyset$.

And there is $H = \{x \in B|Q(x)\}=$ set of all black and white animals that eat shoots and leaves = {pandas, ... and I'm sure something else....}

fleablood
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  • I know that $f(x)$ is the value of $f$ (function) for the value $x$ of its variable. why symbol of property is to form $P(u)$ ? – Almot1960 Dec 15 '18 at 21:23
  • Um... why not be in the form $P(u)$? You can think of $P$ as a function that maps $X \to {YES, NO}$ where $P(x) = YES$ if $x$ has the property and $P(x) = NO$ uf $x$ does not have the property. – fleablood Dec 15 '18 at 22:50
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I give you one exemple :

$P(f,x_0):\forall \varepsilon>0,\exists \eta>0, \forall x\in \mathcal{D}_f\qquad |x-x_0|<\eta \implies |f(x)-f(x_0)|<\varepsilon$

If $f:x\to \lfloor x\rfloor$ and $x_0:=1$ then $P(f,x_0)$ is false, indeed $f$ is not continuous on $x_0=1$

If $f:x\to x^2$ and $x_0:=1$ then $P(f,x_0)$ is true

$f,x_0$ are free variables, $\varepsilon,\eta,x$ are bound variables

Note that we consider $P(u):u\le u^2$

$P(1)$ is true

$\forall u\in \mathbb{R}, P(u)$ is wrong, and $u$ is a bound variable and this proposition doesn't depend on $u$

Stu
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