Can we find a well ordering on an infinite set with no largest element?
$begingroup$
According to the well ordering theorem "Any set can be well ordered". Whenever we have a well ordering on a set, it is not difficult to construct a new well ordering with a largest element.
My question is:
For a given infinite set, can we find a well ordering on it such that there is no largest element?
elementary-set-theory well-orders
edited Jul 2 '17 at 22:52
RQM
1193
asked Jul 2 '17 at 18:53
BumblebeeBumblebee
9,64912551
$endgroup$
|
show 2 more comments
$begingroup$
According to the well ordering theorem "Any set can be well ordered". Whenever we have a well ordering on a set, it is not difficult to construct a new well ordering with a largest element.
My question is:
For a given infinite set, can we find a well ordering on it such that there is no largest element?
elementary-set-theory well-orders
edited Jul 2 '17 at 22:52
RQM
1193
asked Jul 2 '17 at 18:53
BumblebeeBumblebee
9,64912551
$endgroup$
2
$begingroup$
Set of natural numbers.
$endgroup$
– Wuestenfux
Jul 2 '17 at 18:54
2
$begingroup$
Induce the well-order by a bijection with a cardinal number.
$endgroup$
– Daniel Fischer♦
Jul 2 '17 at 18:55
$begingroup$
Use Choice + transfinite induction to define it one at a time?
$endgroup$
– Chappers
Jul 2 '17 at 19:02
$begingroup$
math.stackexchange.com/questions/6501/…
$endgroup$
– Count Iblis
Jul 2 '17 at 19:16
$begingroup$
@CountIblis That question is not related to this one at all.
$endgroup$
– Noah Schweber
Jul 2 '17 at 20:02
|
show 2 more comments
$begingroup$
According to the well ordering theorem "Any set can be well ordered". Whenever we have a well ordering on a set, it is not difficult to construct a new well ordering with a largest element.
My question is:
For a given infinite set, can we find a well ordering on it such that there is no largest element?
elementary-set-theory well-orders
edited Jul 2 '17 at 22:52
RQM
1193
asked Jul 2 '17 at 18:53
BumblebeeBumblebee
9,64912551
$endgroup$
According to the well ordering theorem "Any set can be well ordered". Whenever we have a well ordering on a set, it is not difficult to construct a new well ordering with a largest element.
My question is:
For a given infinite set, can we find a well ordering on it such that there is no largest element?
elementary-set-theory well-orders
elementary-set-theory well-orders
edited Jul 2 '17 at 22:52
RQM
1193
asked Jul 2 '17 at 18:53
BumblebeeBumblebee
9,64912551
edited Jul 2 '17 at 22:52
RQM
1193
asked Jul 2 '17 at 18:53
BumblebeeBumblebee
9,64912551
edited Jul 2 '17 at 22:52
RQM
1193
edited Jul 2 '17 at 22:52
RQM
1193
edited Jul 2 '17 at 22:52
RQM
1193
1193
asked Jul 2 '17 at 18:53
BumblebeeBumblebee
9,64912551
asked Jul 2 '17 at 18:53
BumblebeeBumblebee
9,64912551
asked Jul 2 '17 at 18:53
BumblebeeBumblebee
9,64912551
9,64912551
2
$begingroup$
Set of natural numbers.
$endgroup$
– Wuestenfux
Jul 2 '17 at 18:54
2
$begingroup$
Induce the well-order by a bijection with a cardinal number.
$endgroup$
– Daniel Fischer♦
Jul 2 '17 at 18:55
$begingroup$
Use Choice + transfinite induction to define it one at a time?
$endgroup$
– Chappers
Jul 2 '17 at 19:02
$begingroup$
math.stackexchange.com/questions/6501/…
$endgroup$
– Count Iblis
Jul 2 '17 at 19:16
$begingroup$
@CountIblis That question is not related to this one at all.
$endgroup$
– Noah Schweber
Jul 2 '17 at 20:02
|
show 2 more comments
2
$begingroup$
Set of natural numbers.
$endgroup$
– Wuestenfux
Jul 2 '17 at 18:54
2
$begingroup$
Induce the well-order by a bijection with a cardinal number.
$endgroup$
– Daniel Fischer♦
Jul 2 '17 at 18:55
$begingroup$
Use Choice + transfinite induction to define it one at a time?
$endgroup$
– Chappers
Jul 2 '17 at 19:02
$begingroup$
math.stackexchange.com/questions/6501/…
$endgroup$
– Count Iblis
Jul 2 '17 at 19:16
$begingroup$
@CountIblis That question is not related to this one at all.
$endgroup$
– Noah Schweber
Jul 2 '17 at 20:02
2
2
$begingroup$
Set of natural numbers.
$endgroup$
– Wuestenfux
Jul 2 '17 at 18:54
$begingroup$
Set of natural numbers.
$endgroup$
– Wuestenfux
Jul 2 '17 at 18:54
2
2
$begingroup$
Induce the well-order by a bijection with a cardinal number.
$endgroup$
– Daniel Fischer♦
Jul 2 '17 at 18:55
$begingroup$
Induce the well-order by a bijection with a cardinal number.
$endgroup$
– Daniel Fischer♦
Jul 2 '17 at 18:55
$begingroup$
Use Choice + transfinite induction to define it one at a time?
$endgroup$
– Chappers
Jul 2 '17 at 19:02
$begingroup$
Use Choice + transfinite induction to define it one at a time?
$endgroup$
– Chappers
Jul 2 '17 at 19:02
$begingroup$
math.stackexchange.com/questions/6501/…
$endgroup$
– Count Iblis
Jul 2 '17 at 19:16
$begingroup$
math.stackexchange.com/questions/6501/…
$endgroup$
– Count Iblis
Jul 2 '17 at 19:16
$begingroup$
@CountIblis That question is not related to this one at all.
$endgroup$
– Noah Schweber
Jul 2 '17 at 20:02
$begingroup$
@CountIblis That question is not related to this one at all.
$endgroup$
– Noah Schweber
Jul 2 '17 at 20:02
|
show 2 more comments
3 Answers
3
active
oldest
votes
$begingroup$
Yes. In fact, there is a canonical way to transform a well-ordering into a well-ordering of the same set with no greatest element.
Suppose I have an infinite well-order $X$. Let $F_X$ be the set of elements of $X$ with finitely many things bigger than them - for instance, the greatest element of $X$ (if such an element exists) is in $F_X$, while (if $X$ is infinite) the least element of $X$ (which always exists, since $X$ is a well-ordering) is not in $F_X$.
It's not hard to show that $F_X$ is finite, since otherwise we could build an infinite descending chain in $X$ (exercise). Now let $I_X=Xsetminus F_X$; this is a well-ordered set with no greatest element.
If $X$ is infinite, $I_X$ and $X$ have the same cardinality. If you want an explicit isomorphism, we can "move $F_X$ to the back" - define a new ordering $prec$ on $X$, given by $aprec b$ iff
$a, bin I_X$ and $a<b$ in the original sense of $X$; or
$a, bin F_X$ and $a<b$ in the original sense of $X$; or
$ain F_X, bin I_X$.
answered Jul 2 '17 at 20:05
Noah SchweberNoah Schweber
123k10149284
$endgroup$
add a comment |
$begingroup$
If $A$ is infinite then $A$ can be put in bijection with $A cup mathbb{N}$. Now from the well ordering of $A$ you can ge a well ordering of $A cup mathbb{N}$ without a largest element ( in the obvious way ) and then back, a new well-ordering of $A$ without a largest element.
answered Jul 2 '17 at 19:58
Orest BucicovschiOrest Bucicovschi
28.5k31746
$endgroup$
2
$begingroup$
As an interesting aside, note that for the first sentence, we either need a weak form of the axiom of choice or the assumption (which is present here) that $A$ is well-orderable: it is consistent with ZF that there is an infinite set $A$ which is not in bijection with $Acupmathbb{N}$ (namely, any infinite Dedekind-finite set).
$endgroup$
– Noah Schweber
Jul 2 '17 at 21:48
1
$begingroup$
@Noah Schweber: Very interesting.! To do away with the bijection, I guess we could just move $mathbb{N}$ from the beginning to the end, similarly to the move in your solution.
$endgroup$
– Orest Bucicovschi
Jul 2 '17 at 23:44
add a comment |
$begingroup$
In the usual development of ZFC, a cardinal is defined to the the
least ordinal with a given cardinality. Infinite cardinals are "limit
ordinals" --- as ordered sets they have no largest element.
answered Jul 2 '17 at 18:56
Lord Shark the UnknownLord Shark the Unknown
102k1160132
$endgroup$
add a comment |
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3 Answers
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3 Answers
3
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votes
$begingroup$
Yes. In fact, there is a canonical way to transform a well-ordering into a well-ordering of the same set with no greatest element.
Suppose I have an infinite well-order $X$. Let $F_X$ be the set of elements of $X$ with finitely many things bigger than them - for instance, the greatest element of $X$ (if such an element exists) is in $F_X$, while (if $X$ is infinite) the least element of $X$ (which always exists, since $X$ is a well-ordering) is not in $F_X$.
It's not hard to show that $F_X$ is finite, since otherwise we could build an infinite descending chain in $X$ (exercise). Now let $I_X=Xsetminus F_X$; this is a well-ordered set with no greatest element.
If $X$ is infinite, $I_X$ and $X$ have the same cardinality. If you want an explicit isomorphism, we can "move $F_X$ to the back" - define a new ordering $prec$ on $X$, given by $aprec b$ iff
$a, bin I_X$ and $a<b$ in the original sense of $X$; or
$a, bin F_X$ and $a<b$ in the original sense of $X$; or
$ain F_X, bin I_X$.
answered Jul 2 '17 at 20:05
Noah SchweberNoah Schweber
123k10149284
$endgroup$
add a comment |
$begingroup$
Yes. In fact, there is a canonical way to transform a well-ordering into a well-ordering of the same set with no greatest element.
Suppose I have an infinite well-order $X$. Let $F_X$ be the set of elements of $X$ with finitely many things bigger than them - for instance, the greatest element of $X$ (if such an element exists) is in $F_X$, while (if $X$ is infinite) the least element of $X$ (which always exists, since $X$ is a well-ordering) is not in $F_X$.
It's not hard to show that $F_X$ is finite, since otherwise we could build an infinite descending chain in $X$ (exercise). Now let $I_X=Xsetminus F_X$; this is a well-ordered set with no greatest element.
If $X$ is infinite, $I_X$ and $X$ have the same cardinality. If you want an explicit isomorphism, we can "move $F_X$ to the back" - define a new ordering $prec$ on $X$, given by $aprec b$ iff
$a, bin I_X$ and $a<b$ in the original sense of $X$; or
$a, bin F_X$ and $a<b$ in the original sense of $X$; or
$ain F_X, bin I_X$.
answered Jul 2 '17 at 20:05
Noah SchweberNoah Schweber
123k10149284
$endgroup$
add a comment |
$begingroup$
Yes. In fact, there is a canonical way to transform a well-ordering into a well-ordering of the same set with no greatest element.
Suppose I have an infinite well-order $X$. Let $F_X$ be the set of elements of $X$ with finitely many things bigger than them - for instance, the greatest element of $X$ (if such an element exists) is in $F_X$, while (if $X$ is infinite) the least element of $X$ (which always exists, since $X$ is a well-ordering) is not in $F_X$.
It's not hard to show that $F_X$ is finite, since otherwise we could build an infinite descending chain in $X$ (exercise). Now let $I_X=Xsetminus F_X$; this is a well-ordered set with no greatest element.
If $X$ is infinite, $I_X$ and $X$ have the same cardinality. If you want an explicit isomorphism, we can "move $F_X$ to the back" - define a new ordering $prec$ on $X$, given by $aprec b$ iff
$a, bin I_X$ and $a<b$ in the original sense of $X$; or
$a, bin F_X$ and $a<b$ in the original sense of $X$; or
$ain F_X, bin I_X$.
answered Jul 2 '17 at 20:05
Noah SchweberNoah Schweber
123k10149284
$endgroup$
Yes. In fact, there is a canonical way to transform a well-ordering into a well-ordering of the same set with no greatest element.
Suppose I have an infinite well-order $X$. Let $F_X$ be the set of elements of $X$ with finitely many things bigger than them - for instance, the greatest element of $X$ (if such an element exists) is in $F_X$, while (if $X$ is infinite) the least element of $X$ (which always exists, since $X$ is a well-ordering) is not in $F_X$.
It's not hard to show that $F_X$ is finite, since otherwise we could build an infinite descending chain in $X$ (exercise). Now let $I_X=Xsetminus F_X$; this is a well-ordered set with no greatest element.
If $X$ is infinite, $I_X$ and $X$ have the same cardinality. If you want an explicit isomorphism, we can "move $F_X$ to the back" - define a new ordering $prec$ on $X$, given by $aprec b$ iff
$a, bin I_X$ and $a<b$ in the original sense of $X$; or
$a, bin F_X$ and $a<b$ in the original sense of $X$; or
$ain F_X, bin I_X$.
answered Jul 2 '17 at 20:05
Noah SchweberNoah Schweber
123k10149284
edited Jul 2 '17 at 21:47
answered Jul 2 '17 at 20:05
Noah SchweberNoah Schweber
123k10149284
answered Jul 2 '17 at 20:05
Noah SchweberNoah Schweber
123k10149284
answered Jul 2 '17 at 20:05
Noah SchweberNoah Schweber
123k10149284
123k10149284
add a comment |
add a comment |
$begingroup$
If $A$ is infinite then $A$ can be put in bijection with $A cup mathbb{N}$. Now from the well ordering of $A$ you can ge a well ordering of $A cup mathbb{N}$ without a largest element ( in the obvious way ) and then back, a new well-ordering of $A$ without a largest element.
answered Jul 2 '17 at 19:58
Orest BucicovschiOrest Bucicovschi
28.5k31746
$endgroup$
2
$begingroup$
As an interesting aside, note that for the first sentence, we either need a weak form of the axiom of choice or the assumption (which is present here) that $A$ is well-orderable: it is consistent with ZF that there is an infinite set $A$ which is not in bijection with $Acupmathbb{N}$ (namely, any infinite Dedekind-finite set).
$endgroup$
– Noah Schweber
Jul 2 '17 at 21:48
1
$begingroup$
@Noah Schweber: Very interesting.! To do away with the bijection, I guess we could just move $mathbb{N}$ from the beginning to the end, similarly to the move in your solution.
$endgroup$
– Orest Bucicovschi
Jul 2 '17 at 23:44
add a comment |
$begingroup$
If $A$ is infinite then $A$ can be put in bijection with $A cup mathbb{N}$. Now from the well ordering of $A$ you can ge a well ordering of $A cup mathbb{N}$ without a largest element ( in the obvious way ) and then back, a new well-ordering of $A$ without a largest element.
answered Jul 2 '17 at 19:58
Orest BucicovschiOrest Bucicovschi
28.5k31746
$endgroup$
2
$begingroup$
As an interesting aside, note that for the first sentence, we either need a weak form of the axiom of choice or the assumption (which is present here) that $A$ is well-orderable: it is consistent with ZF that there is an infinite set $A$ which is not in bijection with $Acupmathbb{N}$ (namely, any infinite Dedekind-finite set).
$endgroup$
– Noah Schweber
Jul 2 '17 at 21:48
1
$begingroup$
@Noah Schweber: Very interesting.! To do away with the bijection, I guess we could just move $mathbb{N}$ from the beginning to the end, similarly to the move in your solution.
$endgroup$
– Orest Bucicovschi
Jul 2 '17 at 23:44
add a comment |
$begingroup$
If $A$ is infinite then $A$ can be put in bijection with $A cup mathbb{N}$. Now from the well ordering of $A$ you can ge a well ordering of $A cup mathbb{N}$ without a largest element ( in the obvious way ) and then back, a new well-ordering of $A$ without a largest element.
answered Jul 2 '17 at 19:58
Orest BucicovschiOrest Bucicovschi
28.5k31746
$endgroup$
If $A$ is infinite then $A$ can be put in bijection with $A cup mathbb{N}$. Now from the well ordering of $A$ you can ge a well ordering of $A cup mathbb{N}$ without a largest element ( in the obvious way ) and then back, a new well-ordering of $A$ without a largest element.
answered Jul 2 '17 at 19:58
Orest BucicovschiOrest Bucicovschi
28.5k31746
answered Jul 2 '17 at 19:58
Orest BucicovschiOrest Bucicovschi
28.5k31746
answered Jul 2 '17 at 19:58
Orest BucicovschiOrest Bucicovschi
28.5k31746
answered Jul 2 '17 at 19:58
Orest BucicovschiOrest Bucicovschi
28.5k31746
28.5k31746
2
$begingroup$
As an interesting aside, note that for the first sentence, we either need a weak form of the axiom of choice or the assumption (which is present here) that $A$ is well-orderable: it is consistent with ZF that there is an infinite set $A$ which is not in bijection with $Acupmathbb{N}$ (namely, any infinite Dedekind-finite set).
$endgroup$
– Noah Schweber
Jul 2 '17 at 21:48
1
$begingroup$
@Noah Schweber: Very interesting.! To do away with the bijection, I guess we could just move $mathbb{N}$ from the beginning to the end, similarly to the move in your solution.
$endgroup$
– Orest Bucicovschi
Jul 2 '17 at 23:44
add a comment |
2
$begingroup$
As an interesting aside, note that for the first sentence, we either need a weak form of the axiom of choice or the assumption (which is present here) that $A$ is well-orderable: it is consistent with ZF that there is an infinite set $A$ which is not in bijection with $Acupmathbb{N}$ (namely, any infinite Dedekind-finite set).
$endgroup$
– Noah Schweber
Jul 2 '17 at 21:48
1
$begingroup$
@Noah Schweber: Very interesting.! To do away with the bijection, I guess we could just move $mathbb{N}$ from the beginning to the end, similarly to the move in your solution.
$endgroup$
– Orest Bucicovschi
Jul 2 '17 at 23:44
2
2
$begingroup$
As an interesting aside, note that for the first sentence, we either need a weak form of the axiom of choice or the assumption (which is present here) that $A$ is well-orderable: it is consistent with ZF that there is an infinite set $A$ which is not in bijection with $Acupmathbb{N}$ (namely, any infinite Dedekind-finite set).
$endgroup$
– Noah Schweber
Jul 2 '17 at 21:48
$begingroup$
As an interesting aside, note that for the first sentence, we either need a weak form of the axiom of choice or the assumption (which is present here) that $A$ is well-orderable: it is consistent with ZF that there is an infinite set $A$ which is not in bijection with $Acupmathbb{N}$ (namely, any infinite Dedekind-finite set).
$endgroup$
– Noah Schweber
Jul 2 '17 at 21:48
1
1
$begingroup$
@Noah Schweber: Very interesting.! To do away with the bijection, I guess we could just move $mathbb{N}$ from the beginning to the end, similarly to the move in your solution.
$endgroup$
– Orest Bucicovschi
Jul 2 '17 at 23:44
$begingroup$
@Noah Schweber: Very interesting.! To do away with the bijection, I guess we could just move $mathbb{N}$ from the beginning to the end, similarly to the move in your solution.
$endgroup$
– Orest Bucicovschi
Jul 2 '17 at 23:44
add a comment |
$begingroup$
In the usual development of ZFC, a cardinal is defined to the the
least ordinal with a given cardinality. Infinite cardinals are "limit
ordinals" --- as ordered sets they have no largest element.
answered Jul 2 '17 at 18:56
Lord Shark the UnknownLord Shark the Unknown
102k1160132
$endgroup$
add a comment |
$begingroup$
In the usual development of ZFC, a cardinal is defined to the the
least ordinal with a given cardinality. Infinite cardinals are "limit
ordinals" --- as ordered sets they have no largest element.
answered Jul 2 '17 at 18:56
Lord Shark the UnknownLord Shark the Unknown
102k1160132
$endgroup$
add a comment |
$begingroup$
In the usual development of ZFC, a cardinal is defined to the the
least ordinal with a given cardinality. Infinite cardinals are "limit
ordinals" --- as ordered sets they have no largest element.
answered Jul 2 '17 at 18:56
Lord Shark the UnknownLord Shark the Unknown
102k1160132
$endgroup$
In the usual development of ZFC, a cardinal is defined to the the
least ordinal with a given cardinality. Infinite cardinals are "limit
ordinals" --- as ordered sets they have no largest element.
answered Jul 2 '17 at 18:56
Lord Shark the UnknownLord Shark the Unknown
102k1160132
answered Jul 2 '17 at 18:56
Lord Shark the UnknownLord Shark the Unknown
102k1160132
answered Jul 2 '17 at 18:56
Lord Shark the UnknownLord Shark the Unknown
102k1160132
answered Jul 2 '17 at 18:56
Lord Shark the UnknownLord Shark the Unknown
102k1160132
102k1160132
add a comment |
add a comment |
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$begingroup$
Set of natural numbers.
$endgroup$
– Wuestenfux
Jul 2 '17 at 18:54
2
$begingroup$
Induce the well-order by a bijection with a cardinal number.
$endgroup$
– Daniel Fischer♦
Jul 2 '17 at 18:55
$begingroup$
Use Choice + transfinite induction to define it one at a time?
$endgroup$
– Chappers
Jul 2 '17 at 19:02
$begingroup$
math.stackexchange.com/questions/6501/…
$endgroup$
– Count Iblis
Jul 2 '17 at 19:16
$begingroup$
@CountIblis That question is not related to this one at all.
$endgroup$
– Noah Schweber
Jul 2 '17 at 20:02