Help with AC $Rightarrow$ Zorn's Lemma proof











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I need help with the proof of $AC Rightarrow Zorn$ here http://math.slu.edu/~srivastava/AC.pdf

Specifically the last section of this part:



enter image description here



Using AC, the function $varphi$ is chosen from the set of all chains in $P$, but then for any $S in P$ shouldn't be that $varphi(S) in S$ ? (because is a choice function, I guess). Instead they are being sent to an element outside the chaing (the set $B(S)$)


Thanks in advance










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    up vote
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    down vote

    favorite












    I need help with the proof of $AC Rightarrow Zorn$ here http://math.slu.edu/~srivastava/AC.pdf

    Specifically the last section of this part:



    enter image description here



    Using AC, the function $varphi$ is chosen from the set of all chains in $P$, but then for any $S in P$ shouldn't be that $varphi(S) in S$ ? (because is a choice function, I guess). Instead they are being sent to an element outside the chaing (the set $B(S)$)


    Thanks in advance










    share|cite|improve this question


























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I need help with the proof of $AC Rightarrow Zorn$ here http://math.slu.edu/~srivastava/AC.pdf

      Specifically the last section of this part:



      enter image description here



      Using AC, the function $varphi$ is chosen from the set of all chains in $P$, but then for any $S in P$ shouldn't be that $varphi(S) in S$ ? (because is a choice function, I guess). Instead they are being sent to an element outside the chaing (the set $B(S)$)


      Thanks in advance










      share|cite|improve this question















      I need help with the proof of $AC Rightarrow Zorn$ here http://math.slu.edu/~srivastava/AC.pdf

      Specifically the last section of this part:



      enter image description here



      Using AC, the function $varphi$ is chosen from the set of all chains in $P$, but then for any $S in P$ shouldn't be that $varphi(S) in S$ ? (because is a choice function, I guess). Instead they are being sent to an element outside the chaing (the set $B(S)$)


      Thanks in advance







      logic set-theory axiom-of-choice






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      edited Dec 6 at 2:03

























      asked Dec 6 at 1:38









      mate89

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          Let $C$ be the set of all chains in $P$ which do not contain a maximal element of $P$. Let $X = {B(S)mid Sin C}$ (this is the range of the function $Bcolon Cto X$ defined in the quote).



          And it's this set $X$ that you want to pick a choice function for. (In the quote, it's shown that for all $Sin C$, $B(S)$ is nonempty, so we can do this by AC.)



          Let's call that choice function $f$. Then for all $Sin C$, $f(B(S))in B(S)$. So defining $varphi = fcirc B$, we have $varphi(S)in B(S)$ for all $Sin C$, as desired.






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          • You sir, you just saved the day. Thanks a lot.
            – mate89
            Dec 6 at 1:59











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          1 Answer
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          1 Answer
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          up vote
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          down vote



          accepted










          Let $C$ be the set of all chains in $P$ which do not contain a maximal element of $P$. Let $X = {B(S)mid Sin C}$ (this is the range of the function $Bcolon Cto X$ defined in the quote).



          And it's this set $X$ that you want to pick a choice function for. (In the quote, it's shown that for all $Sin C$, $B(S)$ is nonempty, so we can do this by AC.)



          Let's call that choice function $f$. Then for all $Sin C$, $f(B(S))in B(S)$. So defining $varphi = fcirc B$, we have $varphi(S)in B(S)$ for all $Sin C$, as desired.






          share|cite|improve this answer





















          • You sir, you just saved the day. Thanks a lot.
            – mate89
            Dec 6 at 1:59















          up vote
          1
          down vote



          accepted










          Let $C$ be the set of all chains in $P$ which do not contain a maximal element of $P$. Let $X = {B(S)mid Sin C}$ (this is the range of the function $Bcolon Cto X$ defined in the quote).



          And it's this set $X$ that you want to pick a choice function for. (In the quote, it's shown that for all $Sin C$, $B(S)$ is nonempty, so we can do this by AC.)



          Let's call that choice function $f$. Then for all $Sin C$, $f(B(S))in B(S)$. So defining $varphi = fcirc B$, we have $varphi(S)in B(S)$ for all $Sin C$, as desired.






          share|cite|improve this answer





















          • You sir, you just saved the day. Thanks a lot.
            – mate89
            Dec 6 at 1:59













          up vote
          1
          down vote



          accepted







          up vote
          1
          down vote



          accepted






          Let $C$ be the set of all chains in $P$ which do not contain a maximal element of $P$. Let $X = {B(S)mid Sin C}$ (this is the range of the function $Bcolon Cto X$ defined in the quote).



          And it's this set $X$ that you want to pick a choice function for. (In the quote, it's shown that for all $Sin C$, $B(S)$ is nonempty, so we can do this by AC.)



          Let's call that choice function $f$. Then for all $Sin C$, $f(B(S))in B(S)$. So defining $varphi = fcirc B$, we have $varphi(S)in B(S)$ for all $Sin C$, as desired.






          share|cite|improve this answer












          Let $C$ be the set of all chains in $P$ which do not contain a maximal element of $P$. Let $X = {B(S)mid Sin C}$ (this is the range of the function $Bcolon Cto X$ defined in the quote).



          And it's this set $X$ that you want to pick a choice function for. (In the quote, it's shown that for all $Sin C$, $B(S)$ is nonempty, so we can do this by AC.)



          Let's call that choice function $f$. Then for all $Sin C$, $f(B(S))in B(S)$. So defining $varphi = fcirc B$, we have $varphi(S)in B(S)$ for all $Sin C$, as desired.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 6 at 1:49









          Alex Kruckman

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          26.2k22455












          • You sir, you just saved the day. Thanks a lot.
            – mate89
            Dec 6 at 1:59


















          • You sir, you just saved the day. Thanks a lot.
            – mate89
            Dec 6 at 1:59
















          You sir, you just saved the day. Thanks a lot.
          – mate89
          Dec 6 at 1:59




          You sir, you just saved the day. Thanks a lot.
          – mate89
          Dec 6 at 1:59


















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