what is its definition mathematically of $P(u) $?












0












$begingroup$


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 $P(u ,p_1,p_2,...,p_n)$?










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$endgroup$












  • $begingroup$
    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


















0












$begingroup$


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 $P(u ,p_1,p_2,...,p_n)$?










share|cite|improve this question









$endgroup$












  • $begingroup$
    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
















0












0








0





$begingroup$


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 $P(u ,p_1,p_2,...,p_n)$?










share|cite|improve this question









$endgroup$




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






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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




















  • $begingroup$
    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












4 Answers
4






active

oldest

votes


















1












$begingroup$

$P$ 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 $P$).






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    $p(u)$ 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.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      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



















    1












    $begingroup$

    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....}






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      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



















    1












    $begingroup$

    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 $u$






    share|cite|improve this answer











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      4 Answers
      4






      active

      oldest

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      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      $P$ 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 $P$).






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        $P$ 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 $P$).






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          $P$ 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 $P$).






          share|cite|improve this answer









          $endgroup$



          $P$ 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 $P$).







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 15 '18 at 21:10









          user3482749user3482749

          4,296919




          4,296919























              1












              $begingroup$

              $p(u)$ 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.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                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
















              1












              $begingroup$

              $p(u)$ 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.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                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














              1












              1








              1





              $begingroup$

              $p(u)$ 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.






              share|cite|improve this answer









              $endgroup$



              $p(u)$ 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.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              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


















              • $begingroup$
                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











              1












              $begingroup$

              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....}






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                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
















              1












              $begingroup$

              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....}






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                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














              1












              1








              1





              $begingroup$

              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....}






              share|cite|improve this answer









              $endgroup$



              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....}







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              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


















              • $begingroup$
                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 $P(u)$ ?
              $endgroup$
              – 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












              $begingroup$
              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




              $begingroup$
              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











              1












              $begingroup$

              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 $u$






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                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 $u$






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  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 $u$






                  share|cite|improve this answer











                  $endgroup$



                  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 $u$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 15 '18 at 22:06

























                  answered Dec 15 '18 at 21:43









                  StuStu

                  1,1951414




                  1,1951414






























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