Construction of uncountably many non-isomorphic linear (total) orderings of natural numbers
I would like to find a way to construct uncountably many non-isomorphic linear (total) orderings of natural numbers (as stated in the title).
I've already constructed two non-isomorphic total orderings. They are
$$
1 < 2 < 3 <ldots\
1 > 2 > 3 >ldots
$$
where "$<$" and "$>$" are strict total ordering relations.
Thanks in advance for your help.
elementary-set-theory order-theory natural-numbers
edited Oct 29 '14 at 20:21
Asaf Karagila♦
302k32427757
asked Oct 29 '14 at 20:08
EncorteEncorte
112
add a comment |
I would like to find a way to construct uncountably many non-isomorphic linear (total) orderings of natural numbers (as stated in the title).
I've already constructed two non-isomorphic total orderings. They are
$$
1 < 2 < 3 <ldots\
1 > 2 > 3 >ldots
$$
where "$<$" and "$>$" are strict total ordering relations.
Thanks in advance for your help.
elementary-set-theory order-theory natural-numbers
edited Oct 29 '14 at 20:21
Asaf Karagila♦
302k32427757
asked Oct 29 '14 at 20:08
EncorteEncorte
112
add a comment |
I would like to find a way to construct uncountably many non-isomorphic linear (total) orderings of natural numbers (as stated in the title).
I've already constructed two non-isomorphic total orderings. They are
$$
1 < 2 < 3 <ldots\
1 > 2 > 3 >ldots
$$
where "$<$" and "$>$" are strict total ordering relations.
Thanks in advance for your help.
elementary-set-theory order-theory natural-numbers
edited Oct 29 '14 at 20:21
Asaf Karagila♦
302k32427757
asked Oct 29 '14 at 20:08
EncorteEncorte
112
I would like to find a way to construct uncountably many non-isomorphic linear (total) orderings of natural numbers (as stated in the title).
I've already constructed two non-isomorphic total orderings. They are
$$
1 < 2 < 3 <ldots\
1 > 2 > 3 >ldots
$$
where "$<$" and "$>$" are strict total ordering relations.
Thanks in advance for your help.
elementary-set-theory order-theory natural-numbers
elementary-set-theory order-theory natural-numbers
edited Oct 29 '14 at 20:21
Asaf Karagila♦
302k32427757
asked Oct 29 '14 at 20:08
EncorteEncorte
112
edited Oct 29 '14 at 20:21
Asaf Karagila♦
302k32427757
asked Oct 29 '14 at 20:08
EncorteEncorte
112
edited Oct 29 '14 at 20:21
Asaf Karagila♦
302k32427757
edited Oct 29 '14 at 20:21
Asaf Karagila♦
302k32427757
edited Oct 29 '14 at 20:21
Asaf Karagila♦
302k32427757
302k32427757
asked Oct 29 '14 at 20:08
EncorteEncorte
112
asked Oct 29 '14 at 20:08
EncorteEncorte
112
asked Oct 29 '14 at 20:08
EncorteEncorte
112
112
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
HINT: If you just want "uncountably many", then it's easy to observe that each countable ordinal can be expressed as a linear order of $Bbb N$.
However if you want $2^{aleph_0}$ (which is the maximal number of orders), then given $AsubseteqBbb N$, it is possible to use $A$ to define a unique linear order which looks like copies of the rational numbers with finite intervals between them, whose size depends on $A$.
answered Oct 29 '14 at 20:12
Asaf Karagila♦Asaf Karagila
302k32427757
add a comment |
You can partition $mathbb{N}$ into infinite subsets $S_1$, $S_2$...., and place $S_2$ after $S_1$, $S_3$ after $S_2$ ..... where each $S_i$ is ordered exactly the same way as the natural order or its reverse($<$ or $>$). There are uncountably many partitions of $mathbb{N}$, so there are uncountably many nonisomorphic linear orderings.
answered Nov 24 '18 at 17:50
The Driven manThe Driven man
205
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
HINT: If you just want "uncountably many", then it's easy to observe that each countable ordinal can be expressed as a linear order of $Bbb N$.
However if you want $2^{aleph_0}$ (which is the maximal number of orders), then given $AsubseteqBbb N$, it is possible to use $A$ to define a unique linear order which looks like copies of the rational numbers with finite intervals between them, whose size depends on $A$.
answered Oct 29 '14 at 20:12
Asaf Karagila♦Asaf Karagila
302k32427757
add a comment |
HINT: If you just want "uncountably many", then it's easy to observe that each countable ordinal can be expressed as a linear order of $Bbb N$.
However if you want $2^{aleph_0}$ (which is the maximal number of orders), then given $AsubseteqBbb N$, it is possible to use $A$ to define a unique linear order which looks like copies of the rational numbers with finite intervals between them, whose size depends on $A$.
answered Oct 29 '14 at 20:12
Asaf Karagila♦Asaf Karagila
302k32427757
add a comment |
HINT: If you just want "uncountably many", then it's easy to observe that each countable ordinal can be expressed as a linear order of $Bbb N$.
However if you want $2^{aleph_0}$ (which is the maximal number of orders), then given $AsubseteqBbb N$, it is possible to use $A$ to define a unique linear order which looks like copies of the rational numbers with finite intervals between them, whose size depends on $A$.
answered Oct 29 '14 at 20:12
Asaf Karagila♦Asaf Karagila
302k32427757
HINT: If you just want "uncountably many", then it's easy to observe that each countable ordinal can be expressed as a linear order of $Bbb N$.
However if you want $2^{aleph_0}$ (which is the maximal number of orders), then given $AsubseteqBbb N$, it is possible to use $A$ to define a unique linear order which looks like copies of the rational numbers with finite intervals between them, whose size depends on $A$.
answered Oct 29 '14 at 20:12
Asaf Karagila♦Asaf Karagila
302k32427757
answered Oct 29 '14 at 20:12
Asaf Karagila♦Asaf Karagila
302k32427757
answered Oct 29 '14 at 20:12
Asaf Karagila♦Asaf Karagila
302k32427757
answered Oct 29 '14 at 20:12
Asaf Karagila♦Asaf Karagila
302k32427757
302k32427757
add a comment |
add a comment |
You can partition $mathbb{N}$ into infinite subsets $S_1$, $S_2$...., and place $S_2$ after $S_1$, $S_3$ after $S_2$ ..... where each $S_i$ is ordered exactly the same way as the natural order or its reverse($<$ or $>$). There are uncountably many partitions of $mathbb{N}$, so there are uncountably many nonisomorphic linear orderings.
answered Nov 24 '18 at 17:50
The Driven manThe Driven man
205
add a comment |
You can partition $mathbb{N}$ into infinite subsets $S_1$, $S_2$...., and place $S_2$ after $S_1$, $S_3$ after $S_2$ ..... where each $S_i$ is ordered exactly the same way as the natural order or its reverse($<$ or $>$). There are uncountably many partitions of $mathbb{N}$, so there are uncountably many nonisomorphic linear orderings.
answered Nov 24 '18 at 17:50
The Driven manThe Driven man
205
add a comment |
You can partition $mathbb{N}$ into infinite subsets $S_1$, $S_2$...., and place $S_2$ after $S_1$, $S_3$ after $S_2$ ..... where each $S_i$ is ordered exactly the same way as the natural order or its reverse($<$ or $>$). There are uncountably many partitions of $mathbb{N}$, so there are uncountably many nonisomorphic linear orderings.
answered Nov 24 '18 at 17:50
The Driven manThe Driven man
205
You can partition $mathbb{N}$ into infinite subsets $S_1$, $S_2$...., and place $S_2$ after $S_1$, $S_3$ after $S_2$ ..... where each $S_i$ is ordered exactly the same way as the natural order or its reverse($<$ or $>$). There are uncountably many partitions of $mathbb{N}$, so there are uncountably many nonisomorphic linear orderings.
answered Nov 24 '18 at 17:50
The Driven manThe Driven man
205
answered Nov 24 '18 at 17:50
The Driven manThe Driven man
205
answered Nov 24 '18 at 17:50
The Driven manThe Driven man
205
answered Nov 24 '18 at 17:50
The Driven manThe Driven man
205
205
add a comment |
add a comment |
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