what is its definition mathematically of $P(u) $?
$begingroup$
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 $P(u ,p_1,p_2,...,p_n)$?
set-theory
$endgroup$
add a comment |
$begingroup$
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 $P(u ,p_1,p_2,...,p_n)$?
set-theory
$endgroup$
$begingroup$
math.stackexchange.com/questions/3028366/…
$endgroup$
– Asaf Karagila♦
Dec 16 '18 at 13:01
$begingroup$
Some authors (e.g. K. Kunen) say Comprehension instead of Separation. It depends on how you want to translate the original non-English name.
$endgroup$
– DanielWainfleet
Dec 17 '18 at 7:29
add a comment |
$begingroup$
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 $P(u ,p_1,p_2,...,p_n)$?
set-theory
$endgroup$
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 $P(u ,p_1,p_2,...,p_n)$?
set-theory
set-theory
asked Dec 15 '18 at 21:02
Almot1960Almot1960
2,312824
2,312824
$begingroup$
math.stackexchange.com/questions/3028366/…
$endgroup$
– Asaf Karagila♦
Dec 16 '18 at 13:01
$begingroup$
Some authors (e.g. K. Kunen) say Comprehension instead of Separation. It depends on how you want to translate the original non-English name.
$endgroup$
– DanielWainfleet
Dec 17 '18 at 7:29
add a comment |
$begingroup$
math.stackexchange.com/questions/3028366/…
$endgroup$
– Asaf Karagila♦
Dec 16 '18 at 13:01
$begingroup$
Some authors (e.g. K. Kunen) say Comprehension instead of Separation. It depends on how you want to translate the original non-English name.
$endgroup$
– DanielWainfleet
Dec 17 '18 at 7:29
$begingroup$
math.stackexchange.com/questions/3028366/…
$endgroup$
– Asaf Karagila♦
Dec 16 '18 at 13:01
$begingroup$
math.stackexchange.com/questions/3028366/…
$endgroup$
– Asaf Karagila♦
Dec 16 '18 at 13:01
$begingroup$
Some authors (e.g. K. Kunen) say Comprehension instead of Separation. It depends on how you want to translate the original non-English name.
$endgroup$
– DanielWainfleet
Dec 17 '18 at 7:29
$begingroup$
Some authors (e.g. K. Kunen) say Comprehension instead of Separation. It depends on how you want to translate the original non-English name.
$endgroup$
– DanielWainfleet
Dec 17 '18 at 7:29
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
$P$ 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 $P$).
$endgroup$
add a comment |
$begingroup$
$p(u)$ 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.
$endgroup$
$begingroup$
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 $P(u)$ ?
$endgroup$
– Almot1960
Dec 15 '18 at 21:23
$begingroup$
@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.
$endgroup$
– Noah Schweber
Dec 15 '18 at 21:30
$begingroup$
Thus $P(u)$ not related to the function .and only is a notation . it's true ?
$endgroup$
– Almot1960
Dec 15 '18 at 21:36
$begingroup$
@Almot1960 $p(u)$ is indeed not a function .. it is a property (or predicate).
$endgroup$
– Bram28
Dec 15 '18 at 21:38
add a comment |
$begingroup$
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 $E = {xin A|P(x)} =$ set of all perfect squares that are divisible by $7={49k^2| kin mathbb Z}$.
There is a $F = {x in A|Q(x)} =$ set of all perfect squares that eat shoots and leaves = $emptyset$.
And there is $G = {xin 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....}
$endgroup$
$begingroup$
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 $P(u)$ ?
$endgroup$
– Almot1960
Dec 15 '18 at 21:23
$begingroup$
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.
$endgroup$
– fleablood
Dec 15 '18 at 22:50
add a comment |
$begingroup$
I give you one exemple :
$P(f,x_0):forall varepsilon>0,exists eta>0, forall xin mathcal{D}_fqquad |x-x_0|<eta implies |f(x)-f(x_0)|<varepsilon$
If $f:xto lfloor xrfloor$ and $x_0:=1$ then $P(f,x_0)$ is false, indeed $f$ is not continuous on $x_0=1$
If $f:xto x^2$ 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 $P(u):ule u^2$
$P(1)$ is true
$forall uin mathbb{R}, P(u)$ is wrong, and $u$ is a bound variable and this proposition doesn't depend on $u$
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$P$ 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 $P$).
$endgroup$
add a comment |
$begingroup$
$P$ 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 $P$).
$endgroup$
add a comment |
$begingroup$
$P$ 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 $P$).
$endgroup$
$P$ 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 $P$).
answered Dec 15 '18 at 21:10
user3482749user3482749
4,296919
4,296919
add a comment |
add a comment |
$begingroup$
$p(u)$ 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.
$endgroup$
$begingroup$
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 $P(u)$ ?
$endgroup$
– Almot1960
Dec 15 '18 at 21:23
$begingroup$
@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.
$endgroup$
– Noah Schweber
Dec 15 '18 at 21:30
$begingroup$
Thus $P(u)$ not related to the function .and only is a notation . it's true ?
$endgroup$
– Almot1960
Dec 15 '18 at 21:36
$begingroup$
@Almot1960 $p(u)$ is indeed not a function .. it is a property (or predicate).
$endgroup$
– Bram28
Dec 15 '18 at 21:38
add a comment |
$begingroup$
$p(u)$ 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.
$endgroup$
$begingroup$
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 $P(u)$ ?
$endgroup$
– Almot1960
Dec 15 '18 at 21:23
$begingroup$
@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.
$endgroup$
– Noah Schweber
Dec 15 '18 at 21:30
$begingroup$
Thus $P(u)$ not related to the function .and only is a notation . it's true ?
$endgroup$
– Almot1960
Dec 15 '18 at 21:36
$begingroup$
@Almot1960 $p(u)$ is indeed not a function .. it is a property (or predicate).
$endgroup$
– Bram28
Dec 15 '18 at 21:38
add a comment |
$begingroup$
$p(u)$ 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.
$endgroup$
$p(u)$ 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.
answered Dec 15 '18 at 21:11
Bram28Bram28
63.2k44793
63.2k44793
$begingroup$
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 $P(u)$ ?
$endgroup$
– Almot1960
Dec 15 '18 at 21:23
$begingroup$
@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.
$endgroup$
– Noah Schweber
Dec 15 '18 at 21:30
$begingroup$
Thus $P(u)$ not related to the function .and only is a notation . it's true ?
$endgroup$
– Almot1960
Dec 15 '18 at 21:36
$begingroup$
@Almot1960 $p(u)$ is indeed not a function .. it is a property (or predicate).
$endgroup$
– Bram28
Dec 15 '18 at 21:38
add a comment |
$begingroup$
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 $P(u)$ ?
$endgroup$
– Almot1960
Dec 15 '18 at 21:23
$begingroup$
@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.
$endgroup$
– Noah Schweber
Dec 15 '18 at 21:30
$begingroup$
Thus $P(u)$ not related to the function .and only is a notation . it's true ?
$endgroup$
– Almot1960
Dec 15 '18 at 21:36
$begingroup$
@Almot1960 $p(u)$ is indeed not a function .. it is a property (or predicate).
$endgroup$
– Bram28
Dec 15 '18 at 21:38
$begingroup$
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 $P(u)$ ?
$endgroup$
– Almot1960
Dec 15 '18 at 21:23
$begingroup$
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 $P(u)$ ?
$endgroup$
– Almot1960
Dec 15 '18 at 21:23
$begingroup$
@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.
$endgroup$
– Noah Schweber
Dec 15 '18 at 21:30
$begingroup$
@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.
$endgroup$
– Noah Schweber
Dec 15 '18 at 21:30
$begingroup$
Thus $P(u)$ not related to the function .and only is a notation . it's true ?
$endgroup$
– Almot1960
Dec 15 '18 at 21:36
$begingroup$
Thus $P(u)$ not related to the function .and only is a notation . it's true ?
$endgroup$
– Almot1960
Dec 15 '18 at 21:36
$begingroup$
@Almot1960 $p(u)$ is indeed not a function .. it is a property (or predicate).
$endgroup$
– Bram28
Dec 15 '18 at 21:38
$begingroup$
@Almot1960 $p(u)$ is indeed not a function .. it is a property (or predicate).
$endgroup$
– Bram28
Dec 15 '18 at 21:38
add a comment |
$begingroup$
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 $E = {xin A|P(x)} =$ set of all perfect squares that are divisible by $7={49k^2| kin mathbb Z}$.
There is a $F = {x in A|Q(x)} =$ set of all perfect squares that eat shoots and leaves = $emptyset$.
And there is $G = {xin 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....}
$endgroup$
$begingroup$
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 $P(u)$ ?
$endgroup$
– Almot1960
Dec 15 '18 at 21:23
$begingroup$
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.
$endgroup$
– fleablood
Dec 15 '18 at 22:50
add a comment |
$begingroup$
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 $E = {xin A|P(x)} =$ set of all perfect squares that are divisible by $7={49k^2| kin mathbb Z}$.
There is a $F = {x in A|Q(x)} =$ set of all perfect squares that eat shoots and leaves = $emptyset$.
And there is $G = {xin 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....}
$endgroup$
$begingroup$
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 $P(u)$ ?
$endgroup$
– Almot1960
Dec 15 '18 at 21:23
$begingroup$
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.
$endgroup$
– fleablood
Dec 15 '18 at 22:50
add a comment |
$begingroup$
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 $E = {xin A|P(x)} =$ set of all perfect squares that are divisible by $7={49k^2| kin mathbb Z}$.
There is a $F = {x in A|Q(x)} =$ set of all perfect squares that eat shoots and leaves = $emptyset$.
And there is $G = {xin 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....}
$endgroup$
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 $E = {xin A|P(x)} =$ set of all perfect squares that are divisible by $7={49k^2| kin mathbb Z}$.
There is a $F = {x in A|Q(x)} =$ set of all perfect squares that eat shoots and leaves = $emptyset$.
And there is $G = {xin 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....}
answered Dec 15 '18 at 21:19
fleabloodfleablood
71.9k22687
71.9k22687
$begingroup$
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 $P(u)$ ?
$endgroup$
– Almot1960
Dec 15 '18 at 21:23
$begingroup$
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.
$endgroup$
– fleablood
Dec 15 '18 at 22:50
add a comment |
$begingroup$
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 $P(u)$ ?
$endgroup$
– Almot1960
Dec 15 '18 at 21:23
$begingroup$
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.
$endgroup$
– fleablood
Dec 15 '18 at 22:50
$begingroup$
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 $P(u)$ ?
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– Almot1960
Dec 15 '18 at 21:23
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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 $P(u)$ ?
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– Almot1960
Dec 15 '18 at 21:23
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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.
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– fleablood
Dec 15 '18 at 22:50
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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.
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– fleablood
Dec 15 '18 at 22:50
add a comment |
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I give you one exemple :
$P(f,x_0):forall varepsilon>0,exists eta>0, forall xin mathcal{D}_fqquad |x-x_0|<eta implies |f(x)-f(x_0)|<varepsilon$
If $f:xto lfloor xrfloor$ and $x_0:=1$ then $P(f,x_0)$ is false, indeed $f$ is not continuous on $x_0=1$
If $f:xto x^2$ 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 $P(u):ule u^2$
$P(1)$ is true
$forall uin mathbb{R}, P(u)$ is wrong, and $u$ is a bound variable and this proposition doesn't depend on $u$
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add a comment |
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I give you one exemple :
$P(f,x_0):forall varepsilon>0,exists eta>0, forall xin mathcal{D}_fqquad |x-x_0|<eta implies |f(x)-f(x_0)|<varepsilon$
If $f:xto lfloor xrfloor$ and $x_0:=1$ then $P(f,x_0)$ is false, indeed $f$ is not continuous on $x_0=1$
If $f:xto x^2$ 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 $P(u):ule u^2$
$P(1)$ is true
$forall uin mathbb{R}, P(u)$ is wrong, and $u$ is a bound variable and this proposition doesn't depend on $u$
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add a comment |
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I give you one exemple :
$P(f,x_0):forall varepsilon>0,exists eta>0, forall xin mathcal{D}_fqquad |x-x_0|<eta implies |f(x)-f(x_0)|<varepsilon$
If $f:xto lfloor xrfloor$ and $x_0:=1$ then $P(f,x_0)$ is false, indeed $f$ is not continuous on $x_0=1$
If $f:xto x^2$ 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 $P(u):ule u^2$
$P(1)$ is true
$forall uin mathbb{R}, P(u)$ is wrong, and $u$ is a bound variable and this proposition doesn't depend on $u$
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I give you one exemple :
$P(f,x_0):forall varepsilon>0,exists eta>0, forall xin mathcal{D}_fqquad |x-x_0|<eta implies |f(x)-f(x_0)|<varepsilon$
If $f:xto lfloor xrfloor$ and $x_0:=1$ then $P(f,x_0)$ is false, indeed $f$ is not continuous on $x_0=1$
If $f:xto x^2$ 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 $P(u):ule u^2$
$P(1)$ is true
$forall uin mathbb{R}, P(u)$ is wrong, and $u$ is a bound variable and this proposition doesn't depend on $u$
edited Dec 15 '18 at 22:06
answered Dec 15 '18 at 21:43
StuStu
1,1951414
1,1951414
add a comment |
add a comment |
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math.stackexchange.com/questions/3028366/…
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– Asaf Karagila♦
Dec 16 '18 at 13:01
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Some authors (e.g. K. Kunen) say Comprehension instead of Separation. It depends on how you want to translate the original non-English name.
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– DanielWainfleet
Dec 17 '18 at 7:29