How to prove that this relation is antisymmetric?












1












$begingroup$


Let $F$ be the set of all functions $f : mathbb{R} to mathbb{R}$. A relation $c$ is defined on $F$ by
$f c g$ if and only if $f(x) ≤ g(x)$ for all $x ∈ mathbb{R} $.
Prove that '$c$' is a partial order.



I have proved that the relation is reflexive. Now I must prove that it is antisymmetric (and later transitive).



My working thus far:



Suppose $a, b ∈ F$. We must prove that if $a c b$ and $b c a$ then $a = b$. Now, if $a c b$ and $b c a$ then $f(a) ≤ f(b)$ and $f(b) ≤ f(a)$. Therefore, $f(a) = f(b)$.



Now, how to prove that $a = b$? There is no indication that $f$ is an injective function.






discrete-mathematics relations







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$endgroup$












  • $begingroup$
    The relation is on the set of functions, not the reals. $f$ is a variable!
    $endgroup$
    – ancientmathematician
    Apr 1 '17 at 13:39










  • $begingroup$
    Oops, my mistake everyone...apologies, and sincere thanks to all who helped :)
    $endgroup$
    – Programmer
    Apr 1 '17 at 13:43


















1












$begingroup$


Let $F$ be the set of all functions $f : mathbb{R} to mathbb{R}$. A relation $c$ is defined on $F$ by
$f c g$ if and only if $f(x) ≤ g(x)$ for all $x ∈ mathbb{R} $.
Prove that '$c$' is a partial order.



I have proved that the relation is reflexive. Now I must prove that it is antisymmetric (and later transitive).



My working thus far:



Suppose $a, b ∈ F$. We must prove that if $a c b$ and $b c a$ then $a = b$. Now, if $a c b$ and $b c a$ then $f(a) ≤ f(b)$ and $f(b) ≤ f(a)$. Therefore, $f(a) = f(b)$.



Now, how to prove that $a = b$? There is no indication that $f$ is an injective function.






discrete-mathematics relations







share|cite|improve this question











$endgroup$












  • $begingroup$
    The relation is on the set of functions, not the reals. $f$ is a variable!
    $endgroup$
    – ancientmathematician
    Apr 1 '17 at 13:39










  • $begingroup$
    Oops, my mistake everyone...apologies, and sincere thanks to all who helped :)
    $endgroup$
    – Programmer
    Apr 1 '17 at 13:43
















1












1








1





$begingroup$


Let $F$ be the set of all functions $f : mathbb{R} to mathbb{R}$. A relation $c$ is defined on $F$ by
$f c g$ if and only if $f(x) ≤ g(x)$ for all $x ∈ mathbb{R} $.
Prove that '$c$' is a partial order.



I have proved that the relation is reflexive. Now I must prove that it is antisymmetric (and later transitive).



My working thus far:



Suppose $a, b ∈ F$. We must prove that if $a c b$ and $b c a$ then $a = b$. Now, if $a c b$ and $b c a$ then $f(a) ≤ f(b)$ and $f(b) ≤ f(a)$. Therefore, $f(a) = f(b)$.



Now, how to prove that $a = b$? There is no indication that $f$ is an injective function.






discrete-mathematics relations







share|cite|improve this question











$endgroup$




Let $F$ be the set of all functions $f : mathbb{R} to mathbb{R}$. A relation $c$ is defined on $F$ by
$f c g$ if and only if $f(x) ≤ g(x)$ for all $x ∈ mathbb{R} $.
Prove that '$c$' is a partial order.



I have proved that the relation is reflexive. Now I must prove that it is antisymmetric (and later transitive).



My working thus far:



Suppose $a, b ∈ F$. We must prove that if $a c b$ and $b c a$ then $a = b$. Now, if $a c b$ and $b c a$ then $f(a) ≤ f(b)$ and $f(b) ≤ f(a)$. Therefore, $f(a) = f(b)$.



Now, how to prove that $a = b$? There is no indication that $f$ is an injective function.






discrete-mathematics relations




discrete-mathematics relations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Apr 1 '17 at 14:08









Rafael Wagner

1,8382923




1,8382923










asked Apr 1 '17 at 13:35









ProgrammerProgrammer

409313




409313












  • $begingroup$
    The relation is on the set of functions, not the reals. $f$ is a variable!
    $endgroup$
    – ancientmathematician
    Apr 1 '17 at 13:39










  • $begingroup$
    Oops, my mistake everyone...apologies, and sincere thanks to all who helped :)
    $endgroup$
    – Programmer
    Apr 1 '17 at 13:43




















  • $begingroup$
    The relation is on the set of functions, not the reals. $f$ is a variable!
    $endgroup$
    – ancientmathematician
    Apr 1 '17 at 13:39










  • $begingroup$
    Oops, my mistake everyone...apologies, and sincere thanks to all who helped :)
    $endgroup$
    – Programmer
    Apr 1 '17 at 13:43


















$begingroup$
The relation is on the set of functions, not the reals. $f$ is a variable!
$endgroup$
– ancientmathematician
Apr 1 '17 at 13:39




$begingroup$
The relation is on the set of functions, not the reals. $f$ is a variable!
$endgroup$
– ancientmathematician
Apr 1 '17 at 13:39












$begingroup$
Oops, my mistake everyone...apologies, and sincere thanks to all who helped :)
$endgroup$
– Programmer
Apr 1 '17 at 13:43






$begingroup$
Oops, my mistake everyone...apologies, and sincere thanks to all who helped :)
$endgroup$
– Programmer
Apr 1 '17 at 13:43












3 Answers
3






active

oldest

votes


















3












$begingroup$

I think you are confuse about your own notation. You define the relation as




Let $mathcal{F}(mathbb{R},mathbb{R})$ be the set of all functions $f: mathbb{R}to mathbb{R}$. So define $color{blue}{c}$ as the relation defined on $mathcal{F}(mathbb{R},mathbb{R})$ as, for $f,g in mathcal{F}(mathbb{R},mathbb{R})$ then $f color{blue} c g Leftrightarrow f(x)leq g(x) , forall x in mathbb{R}$.




Now take a look in your proof: You say that $a,b in mathcal{F}(mathbb{R},mathbb{R})$ and then in your proof you use $a,b$ as variables for $f,g$! But your answer is not all wrong. Let's see it



Suppose $a,b in mathcal{F}(mathbb{R},mathbb{R})$. We want to prove that $(a color{blue }c b $ and $b color{blue}c a )$ implies $a = b$. Now what you've done is




  1. $a color{blue}c b implies color{red}forall x in mathbb{R}, a(x)color{red}leq b(x)$

  2. $b color{blue}ca implies color{red}forall x in mathbb{R}, b(x)color{red}leq a(x)$


And then $color{red}{(1)text{ and } (2)text{ together}}$ implies that $forall x in mathbb{R}, a(x) = b(x)$ wich implies that $a = b in mathcal{F}(mathbb{R}, mathbb{R})$. In blue is the relation and in red is important things that should be in your argument. Now you should try to answer transitive. But that should follow directly from arguments of inequality as the one above.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Hi, thanks very much Rafael. Yes, I just realised I confused the notation! :)
    $endgroup$
    – Programmer
    Apr 1 '17 at 13:59










  • $begingroup$
    0/ ok thats fine 0/
    $endgroup$
    – Rafael Wagner
    Apr 1 '17 at 14:01



















1












$begingroup$

The relation is over the set of functions, $F$, and not on the set of real numbers, $mathbb R$.



So, here, your $a$s and $b$s are elements of $F$, i.e., functions that map from $mathbb R$ to $mathbb R$.



In your case, $a c b$ means $a le b$, where $a$ and $b$ are functions in $F$. It does not mean $f(a) le f(b)$.



Hence, showing $f=g$ is sufficient.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Keep in mind that this is a relation on the functions, not the elements. You seem to be getting confused with this difference. If $f prec g$ and $g prec f$ then $f(x) leq g(x)$ and $f(x) geq g(x)$ for all $x in R$. This means $f=g$. No need for injectivity :)






    share|cite|improve this answer









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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      I think you are confuse about your own notation. You define the relation as




      Let $mathcal{F}(mathbb{R},mathbb{R})$ be the set of all functions $f: mathbb{R}to mathbb{R}$. So define $color{blue}{c}$ as the relation defined on $mathcal{F}(mathbb{R},mathbb{R})$ as, for $f,g in mathcal{F}(mathbb{R},mathbb{R})$ then $f color{blue} c g Leftrightarrow f(x)leq g(x) , forall x in mathbb{R}$.




      Now take a look in your proof: You say that $a,b in mathcal{F}(mathbb{R},mathbb{R})$ and then in your proof you use $a,b$ as variables for $f,g$! But your answer is not all wrong. Let's see it



      Suppose $a,b in mathcal{F}(mathbb{R},mathbb{R})$. We want to prove that $(a color{blue }c b $ and $b color{blue}c a )$ implies $a = b$. Now what you've done is




      1. $a color{blue}c b implies color{red}forall x in mathbb{R}, a(x)color{red}leq b(x)$

      2. $b color{blue}ca implies color{red}forall x in mathbb{R}, b(x)color{red}leq a(x)$


      And then $color{red}{(1)text{ and } (2)text{ together}}$ implies that $forall x in mathbb{R}, a(x) = b(x)$ wich implies that $a = b in mathcal{F}(mathbb{R}, mathbb{R})$. In blue is the relation and in red is important things that should be in your argument. Now you should try to answer transitive. But that should follow directly from arguments of inequality as the one above.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Hi, thanks very much Rafael. Yes, I just realised I confused the notation! :)
        $endgroup$
        – Programmer
        Apr 1 '17 at 13:59










      • $begingroup$
        0/ ok thats fine 0/
        $endgroup$
        – Rafael Wagner
        Apr 1 '17 at 14:01
















      3












      $begingroup$

      I think you are confuse about your own notation. You define the relation as




      Let $mathcal{F}(mathbb{R},mathbb{R})$ be the set of all functions $f: mathbb{R}to mathbb{R}$. So define $color{blue}{c}$ as the relation defined on $mathcal{F}(mathbb{R},mathbb{R})$ as, for $f,g in mathcal{F}(mathbb{R},mathbb{R})$ then $f color{blue} c g Leftrightarrow f(x)leq g(x) , forall x in mathbb{R}$.




      Now take a look in your proof: You say that $a,b in mathcal{F}(mathbb{R},mathbb{R})$ and then in your proof you use $a,b$ as variables for $f,g$! But your answer is not all wrong. Let's see it



      Suppose $a,b in mathcal{F}(mathbb{R},mathbb{R})$. We want to prove that $(a color{blue }c b $ and $b color{blue}c a )$ implies $a = b$. Now what you've done is




      1. $a color{blue}c b implies color{red}forall x in mathbb{R}, a(x)color{red}leq b(x)$

      2. $b color{blue}ca implies color{red}forall x in mathbb{R}, b(x)color{red}leq a(x)$


      And then $color{red}{(1)text{ and } (2)text{ together}}$ implies that $forall x in mathbb{R}, a(x) = b(x)$ wich implies that $a = b in mathcal{F}(mathbb{R}, mathbb{R})$. In blue is the relation and in red is important things that should be in your argument. Now you should try to answer transitive. But that should follow directly from arguments of inequality as the one above.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Hi, thanks very much Rafael. Yes, I just realised I confused the notation! :)
        $endgroup$
        – Programmer
        Apr 1 '17 at 13:59










      • $begingroup$
        0/ ok thats fine 0/
        $endgroup$
        – Rafael Wagner
        Apr 1 '17 at 14:01














      3












      3








      3





      $begingroup$

      I think you are confuse about your own notation. You define the relation as




      Let $mathcal{F}(mathbb{R},mathbb{R})$ be the set of all functions $f: mathbb{R}to mathbb{R}$. So define $color{blue}{c}$ as the relation defined on $mathcal{F}(mathbb{R},mathbb{R})$ as, for $f,g in mathcal{F}(mathbb{R},mathbb{R})$ then $f color{blue} c g Leftrightarrow f(x)leq g(x) , forall x in mathbb{R}$.




      Now take a look in your proof: You say that $a,b in mathcal{F}(mathbb{R},mathbb{R})$ and then in your proof you use $a,b$ as variables for $f,g$! But your answer is not all wrong. Let's see it



      Suppose $a,b in mathcal{F}(mathbb{R},mathbb{R})$. We want to prove that $(a color{blue }c b $ and $b color{blue}c a )$ implies $a = b$. Now what you've done is




      1. $a color{blue}c b implies color{red}forall x in mathbb{R}, a(x)color{red}leq b(x)$

      2. $b color{blue}ca implies color{red}forall x in mathbb{R}, b(x)color{red}leq a(x)$


      And then $color{red}{(1)text{ and } (2)text{ together}}$ implies that $forall x in mathbb{R}, a(x) = b(x)$ wich implies that $a = b in mathcal{F}(mathbb{R}, mathbb{R})$. In blue is the relation and in red is important things that should be in your argument. Now you should try to answer transitive. But that should follow directly from arguments of inequality as the one above.






      share|cite|improve this answer











      $endgroup$



      I think you are confuse about your own notation. You define the relation as




      Let $mathcal{F}(mathbb{R},mathbb{R})$ be the set of all functions $f: mathbb{R}to mathbb{R}$. So define $color{blue}{c}$ as the relation defined on $mathcal{F}(mathbb{R},mathbb{R})$ as, for $f,g in mathcal{F}(mathbb{R},mathbb{R})$ then $f color{blue} c g Leftrightarrow f(x)leq g(x) , forall x in mathbb{R}$.




      Now take a look in your proof: You say that $a,b in mathcal{F}(mathbb{R},mathbb{R})$ and then in your proof you use $a,b$ as variables for $f,g$! But your answer is not all wrong. Let's see it



      Suppose $a,b in mathcal{F}(mathbb{R},mathbb{R})$. We want to prove that $(a color{blue }c b $ and $b color{blue}c a )$ implies $a = b$. Now what you've done is




      1. $a color{blue}c b implies color{red}forall x in mathbb{R}, a(x)color{red}leq b(x)$

      2. $b color{blue}ca implies color{red}forall x in mathbb{R}, b(x)color{red}leq a(x)$


      And then $color{red}{(1)text{ and } (2)text{ together}}$ implies that $forall x in mathbb{R}, a(x) = b(x)$ wich implies that $a = b in mathcal{F}(mathbb{R}, mathbb{R})$. In blue is the relation and in red is important things that should be in your argument. Now you should try to answer transitive. But that should follow directly from arguments of inequality as the one above.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Jan 3 at 12:41

























      answered Apr 1 '17 at 13:57









      Rafael WagnerRafael Wagner

      1,8382923




      1,8382923












      • $begingroup$
        Hi, thanks very much Rafael. Yes, I just realised I confused the notation! :)
        $endgroup$
        – Programmer
        Apr 1 '17 at 13:59










      • $begingroup$
        0/ ok thats fine 0/
        $endgroup$
        – Rafael Wagner
        Apr 1 '17 at 14:01


















      • $begingroup$
        Hi, thanks very much Rafael. Yes, I just realised I confused the notation! :)
        $endgroup$
        – Programmer
        Apr 1 '17 at 13:59










      • $begingroup$
        0/ ok thats fine 0/
        $endgroup$
        – Rafael Wagner
        Apr 1 '17 at 14:01
















      $begingroup$
      Hi, thanks very much Rafael. Yes, I just realised I confused the notation! :)
      $endgroup$
      – Programmer
      Apr 1 '17 at 13:59




      $begingroup$
      Hi, thanks very much Rafael. Yes, I just realised I confused the notation! :)
      $endgroup$
      – Programmer
      Apr 1 '17 at 13:59












      $begingroup$
      0/ ok thats fine 0/
      $endgroup$
      – Rafael Wagner
      Apr 1 '17 at 14:01




      $begingroup$
      0/ ok thats fine 0/
      $endgroup$
      – Rafael Wagner
      Apr 1 '17 at 14:01











      1












      $begingroup$

      The relation is over the set of functions, $F$, and not on the set of real numbers, $mathbb R$.



      So, here, your $a$s and $b$s are elements of $F$, i.e., functions that map from $mathbb R$ to $mathbb R$.



      In your case, $a c b$ means $a le b$, where $a$ and $b$ are functions in $F$. It does not mean $f(a) le f(b)$.



      Hence, showing $f=g$ is sufficient.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        The relation is over the set of functions, $F$, and not on the set of real numbers, $mathbb R$.



        So, here, your $a$s and $b$s are elements of $F$, i.e., functions that map from $mathbb R$ to $mathbb R$.



        In your case, $a c b$ means $a le b$, where $a$ and $b$ are functions in $F$. It does not mean $f(a) le f(b)$.



        Hence, showing $f=g$ is sufficient.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          The relation is over the set of functions, $F$, and not on the set of real numbers, $mathbb R$.



          So, here, your $a$s and $b$s are elements of $F$, i.e., functions that map from $mathbb R$ to $mathbb R$.



          In your case, $a c b$ means $a le b$, where $a$ and $b$ are functions in $F$. It does not mean $f(a) le f(b)$.



          Hence, showing $f=g$ is sufficient.






          share|cite|improve this answer









          $endgroup$



          The relation is over the set of functions, $F$, and not on the set of real numbers, $mathbb R$.



          So, here, your $a$s and $b$s are elements of $F$, i.e., functions that map from $mathbb R$ to $mathbb R$.



          In your case, $a c b$ means $a le b$, where $a$ and $b$ are functions in $F$. It does not mean $f(a) le f(b)$.



          Hence, showing $f=g$ is sufficient.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Apr 1 '17 at 13:40









          GoodDeedsGoodDeeds

          10.3k31435




          10.3k31435























              0












              $begingroup$

              Keep in mind that this is a relation on the functions, not the elements. You seem to be getting confused with this difference. If $f prec g$ and $g prec f$ then $f(x) leq g(x)$ and $f(x) geq g(x)$ for all $x in R$. This means $f=g$. No need for injectivity :)






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Keep in mind that this is a relation on the functions, not the elements. You seem to be getting confused with this difference. If $f prec g$ and $g prec f$ then $f(x) leq g(x)$ and $f(x) geq g(x)$ for all $x in R$. This means $f=g$. No need for injectivity :)






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Keep in mind that this is a relation on the functions, not the elements. You seem to be getting confused with this difference. If $f prec g$ and $g prec f$ then $f(x) leq g(x)$ and $f(x) geq g(x)$ for all $x in R$. This means $f=g$. No need for injectivity :)






                  share|cite|improve this answer









                  $endgroup$



                  Keep in mind that this is a relation on the functions, not the elements. You seem to be getting confused with this difference. If $f prec g$ and $g prec f$ then $f(x) leq g(x)$ and $f(x) geq g(x)$ for all $x in R$. This means $f=g$. No need for injectivity :)







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Apr 1 '17 at 13:39









                  phunfdphunfd

                  29629




                  29629






























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                      Cabo Verde

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                        Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º)
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                      Gyllenstierna

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