Mixing
Linearly ordering the power set of a well ordered set with ZF (without AC)
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?
logic set-theory order-theory well-orders
edited Feb 15 at 4:53
Andrés E. Caicedo
65.6k8159250
asked Jan 21 at 6:09
StudentuStudentu
1279
$endgroup$
add a comment |
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?
logic set-theory order-theory well-orders
edited Feb 15 at 4:53
Andrés E. Caicedo
65.6k8159250
asked Jan 21 at 6:09
StudentuStudentu
1279
$endgroup$
1
$begingroup$
Lexicograpohically, by first difference? Where's the snag?
$endgroup$
– bof
Jan 21 at 6:20
add a comment |
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?
logic set-theory order-theory well-orders
edited Feb 15 at 4:53
Andrés E. Caicedo
65.6k8159250
asked Jan 21 at 6:09
StudentuStudentu
1279
$endgroup$
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?
logic set-theory order-theory well-orders
logic set-theory order-theory well-orders
edited Feb 15 at 4:53
Andrés E. Caicedo
65.6k8159250
asked Jan 21 at 6:09
StudentuStudentu
1279
edited Feb 15 at 4:53
Andrés E. Caicedo
65.6k8159250
asked Jan 21 at 6:09
StudentuStudentu
1279
edited Feb 15 at 4:53
Andrés E. Caicedo
65.6k8159250
edited Feb 15 at 4:53
Andrés E. Caicedo
65.6k8159250
edited Feb 15 at 4:53
Andrés E. Caicedo
65.6k8159250
65.6k8159250
asked Jan 21 at 6:09
StudentuStudentu
1279
asked Jan 21 at 6:09
StudentuStudentu
1279
asked Jan 21 at 6:09
StudentuStudentu
1279
1279
1
$begingroup$
Lexicograpohically, by first difference? Where's the snag?
$endgroup$
– bof
Jan 21 at 6:20
add a comment |
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
add a comment |
1 Answer
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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.
answered Jan 21 at 6:20
Ross MillikanRoss Millikan
298k24200374
$endgroup$
$begingroup$
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
|
show 1 more comment
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1 Answer
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active
oldest
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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.
answered Jan 21 at 6:20
Ross MillikanRoss Millikan
298k24200374
$endgroup$
$begingroup$
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
|
show 1 more comment
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.
answered Jan 21 at 6:20
Ross MillikanRoss Millikan
298k24200374
$endgroup$
$begingroup$
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
|
show 1 more comment
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.
answered Jan 21 at 6:20
Ross MillikanRoss Millikan
298k24200374
$endgroup$
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.
answered Jan 21 at 6:20
Ross MillikanRoss Millikan
298k24200374
answered Jan 21 at 6:20
Ross MillikanRoss Millikan
298k24200374
answered Jan 21 at 6:20
Ross MillikanRoss Millikan
298k24200374
answered Jan 21 at 6:20
Ross MillikanRoss Millikan
298k24200374
298k24200374
$begingroup$
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
|
show 1 more comment
$begingroup$
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
|
show 1 more comment
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1
$begingroup$
Lexicograpohically, by first difference? Where's the snag?
$endgroup$
– bof
Jan 21 at 6:20