Linearly ordering the power set of a well ordered set with ZF (without AC)

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As the title says, my question is, how one can use only ZF-theory to prove that the power set of A, whereby (A, <) is a well-ordering, can be linearly ordered?










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  • 1




    $begingroup$
    Lexicograpohically, by first difference? Where's the snag?
    $endgroup$
    – bof
    Jan 21 at 6:20















3












$begingroup$


As the title says, my question is, how one can use only ZF-theory to prove that the power set of A, whereby (A, <) is a well-ordering, can be linearly ordered?










share|cite|improve this question











$endgroup$







  • 1




    $begingroup$
    Lexicograpohically, by first difference? Where's the snag?
    $endgroup$
    – bof
    Jan 21 at 6:20













3












3








3





$begingroup$


As the title says, my question is, how one can use only ZF-theory to prove that the power set of A, whereby (A, <) is a well-ordering, can be linearly ordered?










share|cite|improve this question











$endgroup$




As the title says, my question is, how one can use only ZF-theory to prove that the power set of A, whereby (A, <) is a well-ordering, can be linearly ordered?







elementary-set-theory logic set-theory order-theory well-orders






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edited Jan 21 at 6:15







Studentu

















asked Jan 21 at 6:09









StudentuStudentu

1229




1229







  • 1




    $begingroup$
    Lexicograpohically, by first difference? Where's the snag?
    $endgroup$
    – bof
    Jan 21 at 6:20












  • 1




    $begingroup$
    Lexicograpohically, by first difference? Where's the snag?
    $endgroup$
    – bof
    Jan 21 at 6:20







1




1




$begingroup$
Lexicograpohically, by first difference? Where's the snag?
$endgroup$
– bof
Jan 21 at 6:20




$begingroup$
Lexicograpohically, by first difference? Where's the snag?
$endgroup$
– bof
Jan 21 at 6:20










1 Answer
1






active

oldest

votes


















4












$begingroup$

Find the earliest element in the well order of A where they differ-where it is in one and not the other. Lexicographic order would take the one with the element first.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Not that it matters, but I thought it was usual to represent membership by a $1$ and nonmembership by a $0$, and $0$ comes before $1$.
    $endgroup$
    – bof
    Jan 21 at 6:22










  • $begingroup$
    @bof: I was thinking of each set as a string. If we use the alphabet, a string beginning with ab comes before one beginning with ac because the b is present.
    $endgroup$
    – Ross Millikan
    Jan 21 at 15:11










  • $begingroup$
    @RossMillikan Thanks for your reply! I hve thought of this, as it was suggested in this thread math.stackexchange.com/questions/90078/… but I don't know how to explain that it only uses ZF-theory?
    $endgroup$
    – Studentu
    Jan 21 at 16:28











  • $begingroup$
    Once you have a well order on $A$ (you really just need a total order) you just start down the list. Is the first element in only one subset? If so, that one comes first. Otherwise, keep going. ZF can answer is $x in X$.
    $endgroup$
    – Ross Millikan
    Jan 21 at 16:46










  • $begingroup$
    You don't "just need a total order", you need a well-order. How would you order the power set of $mathbb R$? Which comes first, $mathbb Q$ or $mathbb Rsetminusmathbb Q$? I don't believe you can prove in ZF that there is a total order on the power set of $mathbb R$.
    $endgroup$
    – bof
    Jan 22 at 1:11










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

Find the earliest element in the well order of A where they differ-where it is in one and not the other. Lexicographic order would take the one with the element first.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Not that it matters, but I thought it was usual to represent membership by a $1$ and nonmembership by a $0$, and $0$ comes before $1$.
    $endgroup$
    – bof
    Jan 21 at 6:22










  • $begingroup$
    @bof: I was thinking of each set as a string. If we use the alphabet, a string beginning with ab comes before one beginning with ac because the b is present.
    $endgroup$
    – Ross Millikan
    Jan 21 at 15:11










  • $begingroup$
    @RossMillikan Thanks for your reply! I hve thought of this, as it was suggested in this thread math.stackexchange.com/questions/90078/… but I don't know how to explain that it only uses ZF-theory?
    $endgroup$
    – Studentu
    Jan 21 at 16:28











  • $begingroup$
    Once you have a well order on $A$ (you really just need a total order) you just start down the list. Is the first element in only one subset? If so, that one comes first. Otherwise, keep going. ZF can answer is $x in X$.
    $endgroup$
    – Ross Millikan
    Jan 21 at 16:46










  • $begingroup$
    You don't "just need a total order", you need a well-order. How would you order the power set of $mathbb R$? Which comes first, $mathbb Q$ or $mathbb Rsetminusmathbb Q$? I don't believe you can prove in ZF that there is a total order on the power set of $mathbb R$.
    $endgroup$
    – bof
    Jan 22 at 1:11















4












$begingroup$

Find the earliest element in the well order of A where they differ-where it is in one and not the other. Lexicographic order would take the one with the element first.






share|cite|improve this answer









$endgroup$












  • $begingroup$
    Not that it matters, but I thought it was usual to represent membership by a $1$ and nonmembership by a $0$, and $0$ comes before $1$.
    $endgroup$
    – bof
    Jan 21 at 6:22










  • $begingroup$
    @bof: I was thinking of each set as a string. If we use the alphabet, a string beginning with ab comes before one beginning with ac because the b is present.
    $endgroup$
    – Ross Millikan
    Jan 21 at 15:11










  • $begingroup$
    @RossMillikan Thanks for your reply! I hve thought of this, as it was suggested in this thread math.stackexchange.com/questions/90078/… but I don't know how to explain that it only uses ZF-theory?
    $endgroup$
    – Studentu
    Jan 21 at 16:28











  • $begingroup$
    Once you have a well order on $A$ (you really just need a total order) you just start down the list. Is the first element in only one subset? If so, that one comes first. Otherwise, keep going. ZF can answer is $x in X$.
    $endgroup$
    – Ross Millikan
    Jan 21 at 16:46










  • $begingroup$
    You don't "just need a total order", you need a well-order. How would you order the power set of $mathbb R$? Which comes first, $mathbb Q$ or $mathbb Rsetminusmathbb Q$? I don't believe you can prove in ZF that there is a total order on the power set of $mathbb R$.
    $endgroup$
    – bof
    Jan 22 at 1:11













4












4








4





$begingroup$

Find the earliest element in the well order of A where they differ-where it is in one and not the other. Lexicographic order would take the one with the element first.






share|cite|improve this answer









$endgroup$



Find the earliest element in the well order of A where they differ-where it is in one and not the other. Lexicographic order would take the one with the element first.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 21 at 6:20









Ross MillikanRoss Millikan

295k23198371




295k23198371











  • $begingroup$
    Not that it matters, but I thought it was usual to represent membership by a $1$ and nonmembership by a $0$, and $0$ comes before $1$.
    $endgroup$
    – bof
    Jan 21 at 6:22










  • $begingroup$
    @bof: I was thinking of each set as a string. If we use the alphabet, a string beginning with ab comes before one beginning with ac because the b is present.
    $endgroup$
    – Ross Millikan
    Jan 21 at 15:11










  • $begingroup$
    @RossMillikan Thanks for your reply! I hve thought of this, as it was suggested in this thread math.stackexchange.com/questions/90078/… but I don't know how to explain that it only uses ZF-theory?
    $endgroup$
    – Studentu
    Jan 21 at 16:28











  • $begingroup$
    Once you have a well order on $A$ (you really just need a total order) you just start down the list. Is the first element in only one subset? If so, that one comes first. Otherwise, keep going. ZF can answer is $x in X$.
    $endgroup$
    – Ross Millikan
    Jan 21 at 16:46










  • $begingroup$
    You don't "just need a total order", you need a well-order. How would you order the power set of $mathbb R$? Which comes first, $mathbb Q$ or $mathbb Rsetminusmathbb Q$? I don't believe you can prove in ZF that there is a total order on the power set of $mathbb R$.
    $endgroup$
    – bof
    Jan 22 at 1:11
















  • $begingroup$
    Not that it matters, but I thought it was usual to represent membership by a $1$ and nonmembership by a $0$, and $0$ comes before $1$.
    $endgroup$
    – bof
    Jan 21 at 6:22










  • $begingroup$
    @bof: I was thinking of each set as a string. If we use the alphabet, a string beginning with ab comes before one beginning with ac because the b is present.
    $endgroup$
    – Ross Millikan
    Jan 21 at 15:11










  • $begingroup$
    @RossMillikan Thanks for your reply! I hve thought of this, as it was suggested in this thread math.stackexchange.com/questions/90078/… but I don't know how to explain that it only uses ZF-theory?
    $endgroup$
    – Studentu
    Jan 21 at 16:28











  • $begingroup$
    Once you have a well order on $A$ (you really just need a total order) you just start down the list. Is the first element in only one subset? If so, that one comes first. Otherwise, keep going. ZF can answer is $x in X$.
    $endgroup$
    – Ross Millikan
    Jan 21 at 16:46










  • $begingroup$
    You don't "just need a total order", you need a well-order. How would you order the power set of $mathbb R$? Which comes first, $mathbb Q$ or $mathbb Rsetminusmathbb Q$? I don't believe you can prove in ZF that there is a total order on the power set of $mathbb R$.
    $endgroup$
    – bof
    Jan 22 at 1:11















$begingroup$
Not that it matters, but I thought it was usual to represent membership by a $1$ and nonmembership by a $0$, and $0$ comes before $1$.
$endgroup$
– bof
Jan 21 at 6:22




$begingroup$
Not that it matters, but I thought it was usual to represent membership by a $1$ and nonmembership by a $0$, and $0$ comes before $1$.
$endgroup$
– bof
Jan 21 at 6:22












$begingroup$
@bof: I was thinking of each set as a string. If we use the alphabet, a string beginning with ab comes before one beginning with ac because the b is present.
$endgroup$
– Ross Millikan
Jan 21 at 15:11




$begingroup$
@bof: I was thinking of each set as a string. If we use the alphabet, a string beginning with ab comes before one beginning with ac because the b is present.
$endgroup$
– Ross Millikan
Jan 21 at 15:11












$begingroup$
@RossMillikan Thanks for your reply! I hve thought of this, as it was suggested in this thread math.stackexchange.com/questions/90078/… but I don't know how to explain that it only uses ZF-theory?
$endgroup$
– Studentu
Jan 21 at 16:28





$begingroup$
@RossMillikan Thanks for your reply! I hve thought of this, as it was suggested in this thread math.stackexchange.com/questions/90078/… but I don't know how to explain that it only uses ZF-theory?
$endgroup$
– Studentu
Jan 21 at 16:28













$begingroup$
Once you have a well order on $A$ (you really just need a total order) you just start down the list. Is the first element in only one subset? If so, that one comes first. Otherwise, keep going. ZF can answer is $x in X$.
$endgroup$
– Ross Millikan
Jan 21 at 16:46




$begingroup$
Once you have a well order on $A$ (you really just need a total order) you just start down the list. Is the first element in only one subset? If so, that one comes first. Otherwise, keep going. ZF can answer is $x in X$.
$endgroup$
– Ross Millikan
Jan 21 at 16:46












$begingroup$
You don't "just need a total order", you need a well-order. How would you order the power set of $mathbb R$? Which comes first, $mathbb Q$ or $mathbb Rsetminusmathbb Q$? I don't believe you can prove in ZF that there is a total order on the power set of $mathbb R$.
$endgroup$
– bof
Jan 22 at 1:11




$begingroup$
You don't "just need a total order", you need a well-order. How would you order the power set of $mathbb R$? Which comes first, $mathbb Q$ or $mathbb Rsetminusmathbb Q$? I don't believe you can prove in ZF that there is a total order on the power set of $mathbb R$.
$endgroup$
– bof
Jan 22 at 1:11

















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