Define a bijection such that $f(x)neq x$












0












$begingroup$


Let $A$ a set such that $|A|>1$. Iwant to prove that there exists a bijection $f:Arightarrow A$ such that $f(x)neq x$.



First, by AC, I can obtain a well order for $A$, and w.l.o.g I suppose that it has order type $alpha$. So $A={x_xi}_{xi<alpha}$. Now, define the set $C_xi=Asetminus{x_xi}$. By transfinite recursion I can define the element $y_xi=min C_xisetminus{y_beta:beta<xi}$.



Then, the function $f:Arightarrow A$ such that $f(x_xi)=y_xi$ is the desire function.



Is it fine my approach?






elementary-set-theory







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$endgroup$












  • $begingroup$
    How do you prove that $C_xi setminus {y_beta : beta<xi} neq emptyset$?
    $endgroup$
    – Federico
    Dec 3 '18 at 17:36










  • $begingroup$
    @Federico Fine, this construction works if $alpha$ is a limit ordinal, right?
    $endgroup$
    – Gödel
    Dec 3 '18 at 17:38










  • $begingroup$
    Well, intuitively I would say that it always works, but I don't see immediately why the set is non-empty
    $endgroup$
    – Federico
    Dec 3 '18 at 17:42










  • $begingroup$
    The axiom of choice tag is meant for questions where you're asking specifically about the necessity of choice in the proof. Not for questions whose answers use choice (since that would encompass a significant portion of this site).
    $endgroup$
    – Asaf Karagila♦
    Dec 3 '18 at 17:44






  • 1




    $begingroup$
    @Federico: For example math.stackexchange.com/questions/134152/…
    $endgroup$
    – Asaf Karagila♦
    Dec 4 '18 at 8:08
















0












$begingroup$


Let $A$ a set such that $|A|>1$. Iwant to prove that there exists a bijection $f:Arightarrow A$ such that $f(x)neq x$.



First, by AC, I can obtain a well order for $A$, and w.l.o.g I suppose that it has order type $alpha$. So $A={x_xi}_{xi<alpha}$. Now, define the set $C_xi=Asetminus{x_xi}$. By transfinite recursion I can define the element $y_xi=min C_xisetminus{y_beta:beta<xi}$.



Then, the function $f:Arightarrow A$ such that $f(x_xi)=y_xi$ is the desire function.



Is it fine my approach?






elementary-set-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    How do you prove that $C_xi setminus {y_beta : beta<xi} neq emptyset$?
    $endgroup$
    – Federico
    Dec 3 '18 at 17:36










  • $begingroup$
    @Federico Fine, this construction works if $alpha$ is a limit ordinal, right?
    $endgroup$
    – Gödel
    Dec 3 '18 at 17:38










  • $begingroup$
    Well, intuitively I would say that it always works, but I don't see immediately why the set is non-empty
    $endgroup$
    – Federico
    Dec 3 '18 at 17:42










  • $begingroup$
    The axiom of choice tag is meant for questions where you're asking specifically about the necessity of choice in the proof. Not for questions whose answers use choice (since that would encompass a significant portion of this site).
    $endgroup$
    – Asaf Karagila♦
    Dec 3 '18 at 17:44






  • 1




    $begingroup$
    @Federico: For example math.stackexchange.com/questions/134152/…
    $endgroup$
    – Asaf Karagila♦
    Dec 4 '18 at 8:08














0












0








0





$begingroup$


Let $A$ a set such that $|A|>1$. Iwant to prove that there exists a bijection $f:Arightarrow A$ such that $f(x)neq x$.



First, by AC, I can obtain a well order for $A$, and w.l.o.g I suppose that it has order type $alpha$. So $A={x_xi}_{xi<alpha}$. Now, define the set $C_xi=Asetminus{x_xi}$. By transfinite recursion I can define the element $y_xi=min C_xisetminus{y_beta:beta<xi}$.



Then, the function $f:Arightarrow A$ such that $f(x_xi)=y_xi$ is the desire function.



Is it fine my approach?






elementary-set-theory







share|cite|improve this question











$endgroup$




Let $A$ a set such that $|A|>1$. Iwant to prove that there exists a bijection $f:Arightarrow A$ such that $f(x)neq x$.



First, by AC, I can obtain a well order for $A$, and w.l.o.g I suppose that it has order type $alpha$. So $A={x_xi}_{xi<alpha}$. Now, define the set $C_xi=Asetminus{x_xi}$. By transfinite recursion I can define the element $y_xi=min C_xisetminus{y_beta:beta<xi}$.



Then, the function $f:Arightarrow A$ such that $f(x_xi)=y_xi$ is the desire function.



Is it fine my approach?






elementary-set-theory




elementary-set-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 3 '18 at 18:24









Federico

4,899514




4,899514










asked Dec 3 '18 at 17:31









GödelGödel

1,416319




1,416319












  • $begingroup$
    How do you prove that $C_xi setminus {y_beta : beta<xi} neq emptyset$?
    $endgroup$
    – Federico
    Dec 3 '18 at 17:36










  • $begingroup$
    @Federico Fine, this construction works if $alpha$ is a limit ordinal, right?
    $endgroup$
    – Gödel
    Dec 3 '18 at 17:38










  • $begingroup$
    Well, intuitively I would say that it always works, but I don't see immediately why the set is non-empty
    $endgroup$
    – Federico
    Dec 3 '18 at 17:42










  • $begingroup$
    The axiom of choice tag is meant for questions where you're asking specifically about the necessity of choice in the proof. Not for questions whose answers use choice (since that would encompass a significant portion of this site).
    $endgroup$
    – Asaf Karagila♦
    Dec 3 '18 at 17:44






  • 1




    $begingroup$
    @Federico: For example math.stackexchange.com/questions/134152/…
    $endgroup$
    – Asaf Karagila♦
    Dec 4 '18 at 8:08


















  • $begingroup$
    How do you prove that $C_xi setminus {y_beta : beta<xi} neq emptyset$?
    $endgroup$
    – Federico
    Dec 3 '18 at 17:36










  • $begingroup$
    @Federico Fine, this construction works if $alpha$ is a limit ordinal, right?
    $endgroup$
    – Gödel
    Dec 3 '18 at 17:38










  • $begingroup$
    Well, intuitively I would say that it always works, but I don't see immediately why the set is non-empty
    $endgroup$
    – Federico
    Dec 3 '18 at 17:42










  • $begingroup$
    The axiom of choice tag is meant for questions where you're asking specifically about the necessity of choice in the proof. Not for questions whose answers use choice (since that would encompass a significant portion of this site).
    $endgroup$
    – Asaf Karagila♦
    Dec 3 '18 at 17:44






  • 1




    $begingroup$
    @Federico: For example math.stackexchange.com/questions/134152/…
    $endgroup$
    – Asaf Karagila♦
    Dec 4 '18 at 8:08
















$begingroup$
How do you prove that $C_xi setminus {y_beta : beta<xi} neq emptyset$?
$endgroup$
– Federico
Dec 3 '18 at 17:36




$begingroup$
How do you prove that $C_xi setminus {y_beta : beta<xi} neq emptyset$?
$endgroup$
– Federico
Dec 3 '18 at 17:36












$begingroup$
@Federico Fine, this construction works if $alpha$ is a limit ordinal, right?
$endgroup$
– Gödel
Dec 3 '18 at 17:38




$begingroup$
@Federico Fine, this construction works if $alpha$ is a limit ordinal, right?
$endgroup$
– Gödel
Dec 3 '18 at 17:38












$begingroup$
Well, intuitively I would say that it always works, but I don't see immediately why the set is non-empty
$endgroup$
– Federico
Dec 3 '18 at 17:42




$begingroup$
Well, intuitively I would say that it always works, but I don't see immediately why the set is non-empty
$endgroup$
– Federico
Dec 3 '18 at 17:42












$begingroup$
The axiom of choice tag is meant for questions where you're asking specifically about the necessity of choice in the proof. Not for questions whose answers use choice (since that would encompass a significant portion of this site).
$endgroup$
– Asaf Karagila♦
Dec 3 '18 at 17:44




$begingroup$
The axiom of choice tag is meant for questions where you're asking specifically about the necessity of choice in the proof. Not for questions whose answers use choice (since that would encompass a significant portion of this site).
$endgroup$
– Asaf Karagila♦
Dec 3 '18 at 17:44




1




1




$begingroup$
@Federico: For example math.stackexchange.com/questions/134152/…
$endgroup$
– Asaf Karagila♦
Dec 4 '18 at 8:08




$begingroup$
@Federico: For example math.stackexchange.com/questions/134152/…
$endgroup$
– Asaf Karagila♦
Dec 4 '18 at 8:08










2 Answers
2






active

oldest

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1












$begingroup$

You can also use the fact that every infinite set $X$ can be partitioned into blocks of size 2 using ordinal theory; see this. So we partition $X$



$tag 1 X = cup ; F_lambda text{ with #}(F_lambda) = 2$



Once you have that you are in business. Set



$tag 2 F_{(lambda, rho)} = (F_lambda times F_lambda) setminus {(x,x) , | , x in F_lambda}$



The union of the $F_{(lambda, rho)}$ is a bijective correspondence $f$ with $f(x) ne x$. It can be viewed as a union of transpositions.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Alternatively, you can do the following.



    If $A$ is finite but not a singleton, then it's simple: just cycle the elements.



    If $A$ is infinite, show that $A=A_1 sqcup A_2$ with $|A_1|=|A_2|$ and take any bijection $phi:A_1to A_2$. Then $f|_{A_1}=phi$ and $f|_{A_2}=phi^{-1}$ is a solution.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      $A$ can't be a singleton.
      $endgroup$
      – CopyPasteIt
      Dec 3 '18 at 20:04











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    You can also use the fact that every infinite set $X$ can be partitioned into blocks of size 2 using ordinal theory; see this. So we partition $X$



    $tag 1 X = cup ; F_lambda text{ with #}(F_lambda) = 2$



    Once you have that you are in business. Set



    $tag 2 F_{(lambda, rho)} = (F_lambda times F_lambda) setminus {(x,x) , | , x in F_lambda}$



    The union of the $F_{(lambda, rho)}$ is a bijective correspondence $f$ with $f(x) ne x$. It can be viewed as a union of transpositions.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      You can also use the fact that every infinite set $X$ can be partitioned into blocks of size 2 using ordinal theory; see this. So we partition $X$



      $tag 1 X = cup ; F_lambda text{ with #}(F_lambda) = 2$



      Once you have that you are in business. Set



      $tag 2 F_{(lambda, rho)} = (F_lambda times F_lambda) setminus {(x,x) , | , x in F_lambda}$



      The union of the $F_{(lambda, rho)}$ is a bijective correspondence $f$ with $f(x) ne x$. It can be viewed as a union of transpositions.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        You can also use the fact that every infinite set $X$ can be partitioned into blocks of size 2 using ordinal theory; see this. So we partition $X$



        $tag 1 X = cup ; F_lambda text{ with #}(F_lambda) = 2$



        Once you have that you are in business. Set



        $tag 2 F_{(lambda, rho)} = (F_lambda times F_lambda) setminus {(x,x) , | , x in F_lambda}$



        The union of the $F_{(lambda, rho)}$ is a bijective correspondence $f$ with $f(x) ne x$. It can be viewed as a union of transpositions.






        share|cite|improve this answer









        $endgroup$



        You can also use the fact that every infinite set $X$ can be partitioned into blocks of size 2 using ordinal theory; see this. So we partition $X$



        $tag 1 X = cup ; F_lambda text{ with #}(F_lambda) = 2$



        Once you have that you are in business. Set



        $tag 2 F_{(lambda, rho)} = (F_lambda times F_lambda) setminus {(x,x) , | , x in F_lambda}$



        The union of the $F_{(lambda, rho)}$ is a bijective correspondence $f$ with $f(x) ne x$. It can be viewed as a union of transpositions.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 3 '18 at 20:51









        CopyPasteItCopyPasteIt

        4,1181628




        4,1181628























            1












            $begingroup$

            Alternatively, you can do the following.



            If $A$ is finite but not a singleton, then it's simple: just cycle the elements.



            If $A$ is infinite, show that $A=A_1 sqcup A_2$ with $|A_1|=|A_2|$ and take any bijection $phi:A_1to A_2$. Then $f|_{A_1}=phi$ and $f|_{A_2}=phi^{-1}$ is a solution.






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              $A$ can't be a singleton.
              $endgroup$
              – CopyPasteIt
              Dec 3 '18 at 20:04
















            1












            $begingroup$

            Alternatively, you can do the following.



            If $A$ is finite but not a singleton, then it's simple: just cycle the elements.



            If $A$ is infinite, show that $A=A_1 sqcup A_2$ with $|A_1|=|A_2|$ and take any bijection $phi:A_1to A_2$. Then $f|_{A_1}=phi$ and $f|_{A_2}=phi^{-1}$ is a solution.






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              $A$ can't be a singleton.
              $endgroup$
              – CopyPasteIt
              Dec 3 '18 at 20:04














            1












            1








            1





            $begingroup$

            Alternatively, you can do the following.



            If $A$ is finite but not a singleton, then it's simple: just cycle the elements.



            If $A$ is infinite, show that $A=A_1 sqcup A_2$ with $|A_1|=|A_2|$ and take any bijection $phi:A_1to A_2$. Then $f|_{A_1}=phi$ and $f|_{A_2}=phi^{-1}$ is a solution.






            share|cite|improve this answer











            $endgroup$



            Alternatively, you can do the following.



            If $A$ is finite but not a singleton, then it's simple: just cycle the elements.



            If $A$ is infinite, show that $A=A_1 sqcup A_2$ with $|A_1|=|A_2|$ and take any bijection $phi:A_1to A_2$. Then $f|_{A_1}=phi$ and $f|_{A_2}=phi^{-1}$ is a solution.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 4 '18 at 13:33

























            answered Dec 3 '18 at 17:41









            FedericoFederico

            4,899514




            4,899514








            • 1




              $begingroup$
              $A$ can't be a singleton.
              $endgroup$
              – CopyPasteIt
              Dec 3 '18 at 20:04














            • 1




              $begingroup$
              $A$ can't be a singleton.
              $endgroup$
              – CopyPasteIt
              Dec 3 '18 at 20:04








            1




            1




            $begingroup$
            $A$ can't be a singleton.
            $endgroup$
            – CopyPasteIt
            Dec 3 '18 at 20:04




            $begingroup$
            $A$ can't be a singleton.
            $endgroup$
            – CopyPasteIt
            Dec 3 '18 at 20:04


















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