Proving the Bourbaki–Witt Theorem using Recursion.

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I am trying to prove the Bourbaki–Witt theorem, which states:




Let $left(X,leqright)$ be a partially ordered set and let
$f:Xrightarrow X$ be a function. Suppose that $fleft(xright)geq x$
for all $xin X$ and that every chain in $X$ has a supremum. Then for
all $xin X$, there exists an element $yin Y$ such that $ygeq x$ and
$fleft(yright)=y$.




There is more than one way to prove this theorem. The method I want to use involves Hartogs Lemma (there exists an ordinal which does not inject to a given set) and transfinite recursion. The proofs I have seen start by recursively defining a function as follows:



enter image description here



This is from the following proofwiki article: Bourbaki–Witt theorem
.



First of all, I am uncomfortable with how they defined $g$. It's not rigorous enough for me. My goal is to define $g$ more rigorously/precisely using the recursion theorem. This is the version of the recursion theorem that I will use:




Let $alpha$ be an ordinal, let $Y$ be a non-empty set, and let $ain Y$.
Let $g:Yrightarrow Y$ be a function and let
$h:wpleft(Yright)rightarrow Y$ be a function. Then there exists a
unique function $varphi:alpharightarrow Y$ such that



• $varphileft(emptysetright)=a$,



•
$varphileft(beta^+right)=gleft(varphileft(betaright)right)$
for all $betainalpha$ with $beta^+inalpha$, and



• $varphileft(gammaright)=hleft(varphileft[gammaright]right)$
for every limit ordinal $gammainalpha$.




By $beta^+$, I mean the immediate successor of the ordinal $beta$ and by $wpleft(Yright)$, I mean the power set of $Y$. Also, $varphileft[gammaright]$ is the set $left varphileft(betaright):betaingammaright $.



Is this version of the Recursion Theorem sufficient to define $g$? I know that there are many different versions of the recursion theorem that may be more appropriate than this one. Now using this recursive definition, I will try to define the function $g$ used in the proofwiki article, which I will just label $varphi$:




Let $xin X$ and let $alpha$ be an ordinal such that there does not exist an injection from $alpha$ to $X$. Let
$h:wpleft(Xright)rightarrow X$ be any function such that for all
$Ainwpleft(Xright)$, if $A$ is a chain in $X$, then $hleft(Aright)=sup A$
. By the Recursion Theorem above, there exists
a unique function $varphi:alpharightarrow X$ such that



• $varphileft(emptysetright)=x$,



•
$varphileft(beta^+right)=fleft(varphileft(betaright)right)$
for all $betainalpha$ with $beta^+inalpha$, and



• $varphileft(gammaright)=hleft(varphileft[gammaright]right)$
for every limit ordinal $gammainalpha$.




This is how I started the proof. Now evidently, the function $varphi$ is apparently not the function required. I need to show that if $gammainalpha$ is a limit ordinal, then $varphileft[gammaright]$ is a chain in $X$. That will guarantee that $varphileft[gammaright]$ will have a supremum. I will then have obtained the function in the proofwiki article. This is the part that I am having trouble with. The proofwiki article stated that indeed this is the case and follows from the fact that $fleft(xright)geq x$
for all $xin X$. However, this is not at all obvious to me. How can I show that $varphileft[gammaright]$ is a chain in $X$ for every limit ordinal $gamma$?




set-theory order-theory ordinals fixed-point-theorems




share|cite|improve this question

























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    1
    down vote

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    I am trying to prove the Bourbaki–Witt theorem, which states:




    Let $left(X,leqright)$ be a partially ordered set and let
    $f:Xrightarrow X$ be a function. Suppose that $fleft(xright)geq x$
    for all $xin X$ and that every chain in $X$ has a supremum. Then for
    all $xin X$, there exists an element $yin Y$ such that $ygeq x$ and
    $fleft(yright)=y$.




    There is more than one way to prove this theorem. The method I want to use involves Hartogs Lemma (there exists an ordinal which does not inject to a given set) and transfinite recursion. The proofs I have seen start by recursively defining a function as follows:



    enter image description here



    This is from the following proofwiki article: Bourbaki–Witt theorem
    .



    First of all, I am uncomfortable with how they defined $g$. It's not rigorous enough for me. My goal is to define $g$ more rigorously/precisely using the recursion theorem. This is the version of the recursion theorem that I will use:




    Let $alpha$ be an ordinal, let $Y$ be a non-empty set, and let $ain Y$.
    Let $g:Yrightarrow Y$ be a function and let
    $h:wpleft(Yright)rightarrow Y$ be a function. Then there exists a
    unique function $varphi:alpharightarrow Y$ such that



    • $varphileft(emptysetright)=a$,



    •
    $varphileft(beta^+right)=gleft(varphileft(betaright)right)$
    for all $betainalpha$ with $beta^+inalpha$, and



    • $varphileft(gammaright)=hleft(varphileft[gammaright]right)$
    for every limit ordinal $gammainalpha$.




    By $beta^+$, I mean the immediate successor of the ordinal $beta$ and by $wpleft(Yright)$, I mean the power set of $Y$. Also, $varphileft[gammaright]$ is the set $left varphileft(betaright):betaingammaright $.



    Is this version of the Recursion Theorem sufficient to define $g$? I know that there are many different versions of the recursion theorem that may be more appropriate than this one. Now using this recursive definition, I will try to define the function $g$ used in the proofwiki article, which I will just label $varphi$:




    Let $xin X$ and let $alpha$ be an ordinal such that there does not exist an injection from $alpha$ to $X$. Let
    $h:wpleft(Xright)rightarrow X$ be any function such that for all
    $Ainwpleft(Xright)$, if $A$ is a chain in $X$, then $hleft(Aright)=sup A$
    . By the Recursion Theorem above, there exists
    a unique function $varphi:alpharightarrow X$ such that



    • $varphileft(emptysetright)=x$,



    •
    $varphileft(beta^+right)=fleft(varphileft(betaright)right)$
    for all $betainalpha$ with $beta^+inalpha$, and



    • $varphileft(gammaright)=hleft(varphileft[gammaright]right)$
    for every limit ordinal $gammainalpha$.




    This is how I started the proof. Now evidently, the function $varphi$ is apparently not the function required. I need to show that if $gammainalpha$ is a limit ordinal, then $varphileft[gammaright]$ is a chain in $X$. That will guarantee that $varphileft[gammaright]$ will have a supremum. I will then have obtained the function in the proofwiki article. This is the part that I am having trouble with. The proofwiki article stated that indeed this is the case and follows from the fact that $fleft(xright)geq x$
    for all $xin X$. However, this is not at all obvious to me. How can I show that $varphileft[gammaright]$ is a chain in $X$ for every limit ordinal $gamma$?




    set-theory order-theory ordinals fixed-point-theorems




    share|cite|improve this question























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      I am trying to prove the Bourbaki–Witt theorem, which states:




      Let $left(X,leqright)$ be a partially ordered set and let
      $f:Xrightarrow X$ be a function. Suppose that $fleft(xright)geq x$
      for all $xin X$ and that every chain in $X$ has a supremum. Then for
      all $xin X$, there exists an element $yin Y$ such that $ygeq x$ and
      $fleft(yright)=y$.




      There is more than one way to prove this theorem. The method I want to use involves Hartogs Lemma (there exists an ordinal which does not inject to a given set) and transfinite recursion. The proofs I have seen start by recursively defining a function as follows:



      enter image description here



      This is from the following proofwiki article: Bourbaki–Witt theorem
      .



      First of all, I am uncomfortable with how they defined $g$. It's not rigorous enough for me. My goal is to define $g$ more rigorously/precisely using the recursion theorem. This is the version of the recursion theorem that I will use:




      Let $alpha$ be an ordinal, let $Y$ be a non-empty set, and let $ain Y$.
      Let $g:Yrightarrow Y$ be a function and let
      $h:wpleft(Yright)rightarrow Y$ be a function. Then there exists a
      unique function $varphi:alpharightarrow Y$ such that



      • $varphileft(emptysetright)=a$,



      •
      $varphileft(beta^+right)=gleft(varphileft(betaright)right)$
      for all $betainalpha$ with $beta^+inalpha$, and



      • $varphileft(gammaright)=hleft(varphileft[gammaright]right)$
      for every limit ordinal $gammainalpha$.




      By $beta^+$, I mean the immediate successor of the ordinal $beta$ and by $wpleft(Yright)$, I mean the power set of $Y$. Also, $varphileft[gammaright]$ is the set $left varphileft(betaright):betaingammaright $.



      Is this version of the Recursion Theorem sufficient to define $g$? I know that there are many different versions of the recursion theorem that may be more appropriate than this one. Now using this recursive definition, I will try to define the function $g$ used in the proofwiki article, which I will just label $varphi$:




      Let $xin X$ and let $alpha$ be an ordinal such that there does not exist an injection from $alpha$ to $X$. Let
      $h:wpleft(Xright)rightarrow X$ be any function such that for all
      $Ainwpleft(Xright)$, if $A$ is a chain in $X$, then $hleft(Aright)=sup A$
      . By the Recursion Theorem above, there exists
      a unique function $varphi:alpharightarrow X$ such that



      • $varphileft(emptysetright)=x$,



      •
      $varphileft(beta^+right)=fleft(varphileft(betaright)right)$
      for all $betainalpha$ with $beta^+inalpha$, and



      • $varphileft(gammaright)=hleft(varphileft[gammaright]right)$
      for every limit ordinal $gammainalpha$.




      This is how I started the proof. Now evidently, the function $varphi$ is apparently not the function required. I need to show that if $gammainalpha$ is a limit ordinal, then $varphileft[gammaright]$ is a chain in $X$. That will guarantee that $varphileft[gammaright]$ will have a supremum. I will then have obtained the function in the proofwiki article. This is the part that I am having trouble with. The proofwiki article stated that indeed this is the case and follows from the fact that $fleft(xright)geq x$
      for all $xin X$. However, this is not at all obvious to me. How can I show that $varphileft[gammaright]$ is a chain in $X$ for every limit ordinal $gamma$?




      set-theory order-theory ordinals fixed-point-theorems




      share|cite|improve this question













      I am trying to prove the Bourbaki–Witt theorem, which states:




      Let $left(X,leqright)$ be a partially ordered set and let
      $f:Xrightarrow X$ be a function. Suppose that $fleft(xright)geq x$
      for all $xin X$ and that every chain in $X$ has a supremum. Then for
      all $xin X$, there exists an element $yin Y$ such that $ygeq x$ and
      $fleft(yright)=y$.




      There is more than one way to prove this theorem. The method I want to use involves Hartogs Lemma (there exists an ordinal which does not inject to a given set) and transfinite recursion. The proofs I have seen start by recursively defining a function as follows:



      enter image description here



      This is from the following proofwiki article: Bourbaki–Witt theorem
      .



      First of all, I am uncomfortable with how they defined $g$. It's not rigorous enough for me. My goal is to define $g$ more rigorously/precisely using the recursion theorem. This is the version of the recursion theorem that I will use:




      Let $alpha$ be an ordinal, let $Y$ be a non-empty set, and let $ain Y$.
      Let $g:Yrightarrow Y$ be a function and let
      $h:wpleft(Yright)rightarrow Y$ be a function. Then there exists a
      unique function $varphi:alpharightarrow Y$ such that



      • $varphileft(emptysetright)=a$,



      •
      $varphileft(beta^+right)=gleft(varphileft(betaright)right)$
      for all $betainalpha$ with $beta^+inalpha$, and



      • $varphileft(gammaright)=hleft(varphileft[gammaright]right)$
      for every limit ordinal $gammainalpha$.




      By $beta^+$, I mean the immediate successor of the ordinal $beta$ and by $wpleft(Yright)$, I mean the power set of $Y$. Also, $varphileft[gammaright]$ is the set $left varphileft(betaright):betaingammaright $.



      Is this version of the Recursion Theorem sufficient to define $g$? I know that there are many different versions of the recursion theorem that may be more appropriate than this one. Now using this recursive definition, I will try to define the function $g$ used in the proofwiki article, which I will just label $varphi$:




      Let $xin X$ and let $alpha$ be an ordinal such that there does not exist an injection from $alpha$ to $X$. Let
      $h:wpleft(Xright)rightarrow X$ be any function such that for all
      $Ainwpleft(Xright)$, if $A$ is a chain in $X$, then $hleft(Aright)=sup A$
      . By the Recursion Theorem above, there exists
      a unique function $varphi:alpharightarrow X$ such that



      • $varphileft(emptysetright)=x$,



      •
      $varphileft(beta^+right)=fleft(varphileft(betaright)right)$
      for all $betainalpha$ with $beta^+inalpha$, and



      • $varphileft(gammaright)=hleft(varphileft[gammaright]right)$
      for every limit ordinal $gammainalpha$.




      This is how I started the proof. Now evidently, the function $varphi$ is apparently not the function required. I need to show that if $gammainalpha$ is a limit ordinal, then $varphileft[gammaright]$ is a chain in $X$. That will guarantee that $varphileft[gammaright]$ will have a supremum. I will then have obtained the function in the proofwiki article. This is the part that I am having trouble with. The proofwiki article stated that indeed this is the case and follows from the fact that $fleft(xright)geq x$
      for all $xin X$. However, this is not at all obvious to me. How can I show that $varphileft[gammaright]$ is a chain in $X$ for every limit ordinal $gamma$?





      set-theory order-theory ordinals fixed-point-theorems





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          $f$ is known as inflationary iff for any $x, f(x)geqslant x$. So it is obvious that $f$ inflationary implies $varphi(beta)$ increasing (as mentioned in proofwiki ) for
          $$
          varphi(beta^+)=f(varphi(beta))geqslant varphi(beta)
          $$
          Also $varphileft[gammaright]$ is clearly a chain for $gamma_1<gamma_2$
          $$
          varphi[gamma_1]=varphi(beta):beta<gamma_1 subset varphi(beta):beta<gamma_2=varphi[gamma_2]
          $$
          If $alpha$ is limit ordinal, then for $gamma_1<gamma_2<cdots<alpha$
          $$
          varphi[gamma_1]subsetvarphi[gamma_2]subsetcdotssubsetbigcup_gamma<alphavarphi[gamma]=sup_gamma<alphavarphi[gamma]
          $$
          So we can define $quad h(varphi[gamma])=sup_gamma<alphavarphi[gamma]$.






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            $f$ is known as inflationary iff for any $x, f(x)geqslant x$. So it is obvious that $f$ inflationary implies $varphi(beta)$ increasing (as mentioned in proofwiki ) for
            $$
            varphi(beta^+)=f(varphi(beta))geqslant varphi(beta)
            $$
            Also $varphileft[gammaright]$ is clearly a chain for $gamma_1<gamma_2$
            $$
            varphi[gamma_1]=varphi(beta):beta<gamma_1 subset varphi(beta):beta<gamma_2=varphi[gamma_2]
            $$
            If $alpha$ is limit ordinal, then for $gamma_1<gamma_2<cdots<alpha$
            $$
            varphi[gamma_1]subsetvarphi[gamma_2]subsetcdotssubsetbigcup_gamma<alphavarphi[gamma]=sup_gamma<alphavarphi[gamma]
            $$
            So we can define $quad h(varphi[gamma])=sup_gamma<alphavarphi[gamma]$.






            share|cite|improve this answer



























              up vote
              0
              down vote













              $f$ is known as inflationary iff for any $x, f(x)geqslant x$. So it is obvious that $f$ inflationary implies $varphi(beta)$ increasing (as mentioned in proofwiki ) for
              $$
              varphi(beta^+)=f(varphi(beta))geqslant varphi(beta)
              $$
              Also $varphileft[gammaright]$ is clearly a chain for $gamma_1<gamma_2$
              $$
              varphi[gamma_1]=varphi(beta):beta<gamma_1 subset varphi(beta):beta<gamma_2=varphi[gamma_2]
              $$
              If $alpha$ is limit ordinal, then for $gamma_1<gamma_2<cdots<alpha$
              $$
              varphi[gamma_1]subsetvarphi[gamma_2]subsetcdotssubsetbigcup_gamma<alphavarphi[gamma]=sup_gamma<alphavarphi[gamma]
              $$
              So we can define $quad h(varphi[gamma])=sup_gamma<alphavarphi[gamma]$.






              share|cite|improve this answer

























                up vote
                0
                down vote










                up vote
                0
                down vote









                $f$ is known as inflationary iff for any $x, f(x)geqslant x$. So it is obvious that $f$ inflationary implies $varphi(beta)$ increasing (as mentioned in proofwiki ) for
                $$
                varphi(beta^+)=f(varphi(beta))geqslant varphi(beta)
                $$
                Also $varphileft[gammaright]$ is clearly a chain for $gamma_1<gamma_2$
                $$
                varphi[gamma_1]=varphi(beta):beta<gamma_1 subset varphi(beta):beta<gamma_2=varphi[gamma_2]
                $$
                If $alpha$ is limit ordinal, then for $gamma_1<gamma_2<cdots<alpha$
                $$
                varphi[gamma_1]subsetvarphi[gamma_2]subsetcdotssubsetbigcup_gamma<alphavarphi[gamma]=sup_gamma<alphavarphi[gamma]
                $$
                So we can define $quad h(varphi[gamma])=sup_gamma<alphavarphi[gamma]$.






                share|cite|improve this answer















                $f$ is known as inflationary iff for any $x, f(x)geqslant x$. So it is obvious that $f$ inflationary implies $varphi(beta)$ increasing (as mentioned in proofwiki ) for
                $$
                varphi(beta^+)=f(varphi(beta))geqslant varphi(beta)
                $$
                Also $varphileft[gammaright]$ is clearly a chain for $gamma_1<gamma_2$
                $$
                varphi[gamma_1]=varphi(beta):beta<gamma_1 subset varphi(beta):beta<gamma_2=varphi[gamma_2]
                $$
                If $alpha$ is limit ordinal, then for $gamma_1<gamma_2<cdots<alpha$
                $$
                varphi[gamma_1]subsetvarphi[gamma_2]subsetcdotssubsetbigcup_gamma<alphavarphi[gamma]=sup_gamma<alphavarphi[gamma]
                $$
                So we can define $quad h(varphi[gamma])=sup_gamma<alphavarphi[gamma]$.







                share|cite|improve this answer















                share|cite|improve this answer



                share|cite|improve this answer








                edited 2 days ago


























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                Math Wizard

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