Mixing
Well order of naturals [closed]
$begingroup$
I have an exercise that asks me for 15 non-isomorphic well order types of natural numbers, I have some, can you help me with other well uncommon orders?
Thank you
elementary-set-theory order-theory well-orders
asked Jan 22 at 1:01
Lennis MarianaLennis Mariana
255
$endgroup$
closed as off-topic by Eevee Trainer, Lord Shark the Unknown, mrtaurho, José Carlos Santos, Chris Custer Jan 23 at 8:59
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Lord Shark the Unknown, mrtaurho, José Carlos Santos, Chris Custer
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I have an exercise that asks me for 15 non-isomorphic well order types of natural numbers, I have some, can you help me with other well uncommon orders?
Thank you
elementary-set-theory order-theory well-orders
asked Jan 22 at 1:01
Lennis MarianaLennis Mariana
255
$endgroup$
closed as off-topic by Eevee Trainer, Lord Shark the Unknown, mrtaurho, José Carlos Santos, Chris Custer Jan 23 at 8:59
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Lord Shark the Unknown, mrtaurho, José Carlos Santos, Chris Custer
If this question can be reworded to fit the rules in the help center, please edit the question.
8
$begingroup$
Suggest you put in question which well orders you already found. This will show you did some work on it, which the site usually wants to see or they put questions "on hold".
$endgroup$
– coffeemath
Jan 22 at 1:06
1
$begingroup$
I hate to say this ordering them so that $a < b$ if the either the remainder of $a$ when divided by $k$ is less than $b$'s remainder or if their remainders are the same if $a$ is smaller than $b$ in the usual order; you can do that for $15$ different values of $k$. They aren't isomorphic but that really seems to defeat the spirit of the question.
$endgroup$
– fleablood
Jan 22 at 1:21
1
$begingroup$
Take as many countable well-orders as you like, from the countable ordinals. $omega,omega+1, ldots, omega + 14$ are already $15$ different ones. Transport them to the natural numbers via a bijection.
$endgroup$
– Henno Brandsma
Jan 22 at 6:52
$begingroup$
15 is a joke... I know uncountably many :)
$endgroup$
– Jonathan
Jan 22 at 12:09
add a comment |
$begingroup$
I have an exercise that asks me for 15 non-isomorphic well order types of natural numbers, I have some, can you help me with other well uncommon orders?
Thank you
elementary-set-theory order-theory well-orders
asked Jan 22 at 1:01
Lennis MarianaLennis Mariana
255
$endgroup$
I have an exercise that asks me for 15 non-isomorphic well order types of natural numbers, I have some, can you help me with other well uncommon orders?
Thank you
elementary-set-theory order-theory well-orders
elementary-set-theory order-theory well-orders
asked Jan 22 at 1:01
Lennis MarianaLennis Mariana
255
asked Jan 22 at 1:01
Lennis MarianaLennis Mariana
255
asked Jan 22 at 1:01
Lennis MarianaLennis Mariana
255
asked Jan 22 at 1:01
Lennis MarianaLennis Mariana
255
asked Jan 22 at 1:01
Lennis MarianaLennis Mariana
255
255
closed as off-topic by Eevee Trainer, Lord Shark the Unknown, mrtaurho, José Carlos Santos, Chris Custer Jan 23 at 8:59
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Lord Shark the Unknown, mrtaurho, José Carlos Santos, Chris Custer
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Eevee Trainer, Lord Shark the Unknown, mrtaurho, José Carlos Santos, Chris Custer Jan 23 at 8:59
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, Lord Shark the Unknown, mrtaurho, José Carlos Santos, Chris Custer
If this question can be reworded to fit the rules in the help center, please edit the question.
8
$begingroup$
Suggest you put in question which well orders you already found. This will show you did some work on it, which the site usually wants to see or they put questions "on hold".
$endgroup$
– coffeemath
Jan 22 at 1:06
1
$begingroup$
I hate to say this ordering them so that $a < b$ if the either the remainder of $a$ when divided by $k$ is less than $b$'s remainder or if their remainders are the same if $a$ is smaller than $b$ in the usual order; you can do that for $15$ different values of $k$. They aren't isomorphic but that really seems to defeat the spirit of the question.
$endgroup$
– fleablood
Jan 22 at 1:21
1
$begingroup$
Take as many countable well-orders as you like, from the countable ordinals. $omega,omega+1, ldots, omega + 14$ are already $15$ different ones. Transport them to the natural numbers via a bijection.
$endgroup$
– Henno Brandsma
Jan 22 at 6:52
$begingroup$
15 is a joke... I know uncountably many :)
$endgroup$
– Jonathan
Jan 22 at 12:09
add a comment |
8
$begingroup$
Suggest you put in question which well orders you already found. This will show you did some work on it, which the site usually wants to see or they put questions "on hold".
$endgroup$
– coffeemath
Jan 22 at 1:06
1
$begingroup$
I hate to say this ordering them so that $a < b$ if the either the remainder of $a$ when divided by $k$ is less than $b$'s remainder or if their remainders are the same if $a$ is smaller than $b$ in the usual order; you can do that for $15$ different values of $k$. They aren't isomorphic but that really seems to defeat the spirit of the question.
$endgroup$
– fleablood
Jan 22 at 1:21
1
$begingroup$
Take as many countable well-orders as you like, from the countable ordinals. $omega,omega+1, ldots, omega + 14$ are already $15$ different ones. Transport them to the natural numbers via a bijection.
$endgroup$
– Henno Brandsma
Jan 22 at 6:52
$begingroup$
15 is a joke... I know uncountably many :)
$endgroup$
– Jonathan
Jan 22 at 12:09
8
8
$begingroup$
Suggest you put in question which well orders you already found. This will show you did some work on it, which the site usually wants to see or they put questions "on hold".
$endgroup$
– coffeemath
Jan 22 at 1:06
$begingroup$
Suggest you put in question which well orders you already found. This will show you did some work on it, which the site usually wants to see or they put questions "on hold".
$endgroup$
– coffeemath
Jan 22 at 1:06
1
1
$begingroup$
I hate to say this ordering them so that $a < b$ if the either the remainder of $a$ when divided by $k$ is less than $b$'s remainder or if their remainders are the same if $a$ is smaller than $b$ in the usual order; you can do that for $15$ different values of $k$. They aren't isomorphic but that really seems to defeat the spirit of the question.
$endgroup$
– fleablood
Jan 22 at 1:21
$begingroup$
I hate to say this ordering them so that $a < b$ if the either the remainder of $a$ when divided by $k$ is less than $b$'s remainder or if their remainders are the same if $a$ is smaller than $b$ in the usual order; you can do that for $15$ different values of $k$. They aren't isomorphic but that really seems to defeat the spirit of the question.
$endgroup$
– fleablood
Jan 22 at 1:21
1
1
$begingroup$
Take as many countable well-orders as you like, from the countable ordinals. $omega,omega+1, ldots, omega + 14$ are already $15$ different ones. Transport them to the natural numbers via a bijection.
$endgroup$
– Henno Brandsma
Jan 22 at 6:52
$begingroup$
Take as many countable well-orders as you like, from the countable ordinals. $omega,omega+1, ldots, omega + 14$ are already $15$ different ones. Transport them to the natural numbers via a bijection.
$endgroup$
– Henno Brandsma
Jan 22 at 6:52
$begingroup$
15 is a joke... I know uncountably many :)
$endgroup$
– Jonathan
Jan 22 at 12:09
$begingroup$
15 is a joke... I know uncountably many :)
$endgroup$
– Jonathan
Jan 22 at 12:09
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$omega_0$ + n for all n in N are equinumerous to N
well ordered and not pairwise order isomorphic.
1,2,3,... 0; 2,3,4,... 0,1; and so on.
$omega_0$ + $omega_0$ + n is yet another bunch of examples.
For example placing all of the even numbers in order before all of the odd numbers in order. 0,2,3,... 1,3,5,...
This idea can be continued indefinitely
0,3,6,... 1,4,7,... 2,5.8,...
and so forth for other sets of integers modulus n.
But wait, there are scads more based upon even larger ordinals such as $omega_0 × omega_0$. Here's a small wow
0 2 5 ...
1 4 ...
3 ...
•••
For bigger wows, go for higher powers of $omega_0$.
Beyond all of those are colossal wowies.
answered Jan 22 at 4:17
William ElliotWilliam Elliot
8,5922720
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$omega_0$ + n for all n in N are equinumerous to N
well ordered and not pairwise order isomorphic.
1,2,3,... 0; 2,3,4,... 0,1; and so on.
$omega_0$ + $omega_0$ + n is yet another bunch of examples.
For example placing all of the even numbers in order before all of the odd numbers in order. 0,2,3,... 1,3,5,...
This idea can be continued indefinitely
0,3,6,... 1,4,7,... 2,5.8,...
and so forth for other sets of integers modulus n.
But wait, there are scads more based upon even larger ordinals such as $omega_0 × omega_0$. Here's a small wow
0 2 5 ...
1 4 ...
3 ...
•••
For bigger wows, go for higher powers of $omega_0$.
Beyond all of those are colossal wowies.
answered Jan 22 at 4:17
William ElliotWilliam Elliot
8,5922720
$endgroup$
add a comment |
$begingroup$
$omega_0$ + n for all n in N are equinumerous to N
well ordered and not pairwise order isomorphic.
1,2,3,... 0; 2,3,4,... 0,1; and so on.
$omega_0$ + $omega_0$ + n is yet another bunch of examples.
For example placing all of the even numbers in order before all of the odd numbers in order. 0,2,3,... 1,3,5,...
This idea can be continued indefinitely
0,3,6,... 1,4,7,... 2,5.8,...
and so forth for other sets of integers modulus n.
But wait, there are scads more based upon even larger ordinals such as $omega_0 × omega_0$. Here's a small wow
0 2 5 ...
1 4 ...
3 ...
•••
For bigger wows, go for higher powers of $omega_0$.
Beyond all of those are colossal wowies.
answered Jan 22 at 4:17
William ElliotWilliam Elliot
8,5922720
$endgroup$
add a comment |
$begingroup$
$omega_0$ + n for all n in N are equinumerous to N
well ordered and not pairwise order isomorphic.
1,2,3,... 0; 2,3,4,... 0,1; and so on.
$omega_0$ + $omega_0$ + n is yet another bunch of examples.
For example placing all of the even numbers in order before all of the odd numbers in order. 0,2,3,... 1,3,5,...
This idea can be continued indefinitely
0,3,6,... 1,4,7,... 2,5.8,...
and so forth for other sets of integers modulus n.
But wait, there are scads more based upon even larger ordinals such as $omega_0 × omega_0$. Here's a small wow
0 2 5 ...
1 4 ...
3 ...
•••
For bigger wows, go for higher powers of $omega_0$.
Beyond all of those are colossal wowies.
answered Jan 22 at 4:17
William ElliotWilliam Elliot
8,5922720
$endgroup$
$omega_0$ + n for all n in N are equinumerous to N
well ordered and not pairwise order isomorphic.
1,2,3,... 0; 2,3,4,... 0,1; and so on.
$omega_0$ + $omega_0$ + n is yet another bunch of examples.
For example placing all of the even numbers in order before all of the odd numbers in order. 0,2,3,... 1,3,5,...
This idea can be continued indefinitely
0,3,6,... 1,4,7,... 2,5.8,...
and so forth for other sets of integers modulus n.
But wait, there are scads more based upon even larger ordinals such as $omega_0 × omega_0$. Here's a small wow
0 2 5 ...
1 4 ...
3 ...
•••
For bigger wows, go for higher powers of $omega_0$.
Beyond all of those are colossal wowies.
answered Jan 22 at 4:17
William ElliotWilliam Elliot
8,5922720
answered Jan 22 at 4:17
William ElliotWilliam Elliot
8,5922720
answered Jan 22 at 4:17
William ElliotWilliam Elliot
8,5922720
answered Jan 22 at 4:17
William ElliotWilliam Elliot
8,5922720
8,5922720
add a comment |
add a comment |
8
$begingroup$
Suggest you put in question which well orders you already found. This will show you did some work on it, which the site usually wants to see or they put questions "on hold".
$endgroup$
– coffeemath
Jan 22 at 1:06
1
$begingroup$
I hate to say this ordering them so that $a < b$ if the either the remainder of $a$ when divided by $k$ is less than $b$'s remainder or if their remainders are the same if $a$ is smaller than $b$ in the usual order; you can do that for $15$ different values of $k$. They aren't isomorphic but that really seems to defeat the spirit of the question.
$endgroup$
– fleablood
Jan 22 at 1:21
1
$begingroup$
Take as many countable well-orders as you like, from the countable ordinals. $omega,omega+1, ldots, omega + 14$ are already $15$ different ones. Transport them to the natural numbers via a bijection.
$endgroup$
– Henno Brandsma
Jan 22 at 6:52
$begingroup$
15 is a joke... I know uncountably many :)
$endgroup$
– Jonathan
Jan 22 at 12:09