Existence of a countable family of pairwise non-similar partial orders with $2^c$ maximal elements, and no...
up vote
1
down vote
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Give an example (if exists) for a countable family of partially ordered sets such that:
a) they are non-similar in pairs (explained under the question)
b) each set has exactly $2^c$ maximal elements and $0$ minimal elements.
If such family of sets doesn't exist, prove it.
Two partially ordered sets are similar if there exists a similarity function $f$. Similarity function is a bijection such that $f$ and $f^{-1}$ order-preserving.
I seriously can't even conclude whether such family exists, so I would appreciate any help! Thanks in advance!
EDIT: I forgot to mention that there needs to be $textbf{exactly}$ $2^c$ maximal elements.
order-theory
asked Nov 22 at 16:44
mathbbandstuff
251111
add a comment |
up vote
1
down vote
favorite
Give an example (if exists) for a countable family of partially ordered sets such that:
a) they are non-similar in pairs (explained under the question)
b) each set has exactly $2^c$ maximal elements and $0$ minimal elements.
If such family of sets doesn't exist, prove it.
Two partially ordered sets are similar if there exists a similarity function $f$. Similarity function is a bijection such that $f$ and $f^{-1}$ order-preserving.
I seriously can't even conclude whether such family exists, so I would appreciate any help! Thanks in advance!
EDIT: I forgot to mention that there needs to be $textbf{exactly}$ $2^c$ maximal elements.
order-theory
asked Nov 22 at 16:44
mathbbandstuff
251111
For an easier problem, how about if you instead wanted there to be $0$ maximal elements and $0$ minimal elements?
– Eric Wofsey
Nov 22 at 16:53
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Give an example (if exists) for a countable family of partially ordered sets such that:
a) they are non-similar in pairs (explained under the question)
b) each set has exactly $2^c$ maximal elements and $0$ minimal elements.
If such family of sets doesn't exist, prove it.
Two partially ordered sets are similar if there exists a similarity function $f$. Similarity function is a bijection such that $f$ and $f^{-1}$ order-preserving.
I seriously can't even conclude whether such family exists, so I would appreciate any help! Thanks in advance!
EDIT: I forgot to mention that there needs to be $textbf{exactly}$ $2^c$ maximal elements.
order-theory
asked Nov 22 at 16:44
mathbbandstuff
251111
Give an example (if exists) for a countable family of partially ordered sets such that:
a) they are non-similar in pairs (explained under the question)
b) each set has exactly $2^c$ maximal elements and $0$ minimal elements.
If such family of sets doesn't exist, prove it.
Two partially ordered sets are similar if there exists a similarity function $f$. Similarity function is a bijection such that $f$ and $f^{-1}$ order-preserving.
I seriously can't even conclude whether such family exists, so I would appreciate any help! Thanks in advance!
EDIT: I forgot to mention that there needs to be $textbf{exactly}$ $2^c$ maximal elements.
order-theory
order-theory
asked Nov 22 at 16:44
mathbbandstuff
251111
asked Nov 22 at 16:44
mathbbandstuff
251111
edited Nov 22 at 17:08
asked Nov 22 at 16:44
mathbbandstuff
251111
asked Nov 22 at 16:44
mathbbandstuff
251111
asked Nov 22 at 16:44
mathbbandstuff
251111
251111
For an easier problem, how about if you instead wanted there to be $0$ maximal elements and $0$ minimal elements?
– Eric Wofsey
Nov 22 at 16:53
add a comment |
For an easier problem, how about if you instead wanted there to be $0$ maximal elements and $0$ minimal elements?
– Eric Wofsey
Nov 22 at 16:53
For an easier problem, how about if you instead wanted there to be $0$ maximal elements and $0$ minimal elements?
– Eric Wofsey
Nov 22 at 16:53
For an easier problem, how about if you instead wanted there to be $0$ maximal elements and $0$ minimal elements?
– Eric Wofsey
Nov 22 at 16:53
add a comment |
1 Answer
1
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oldest
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up vote
3
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Consider a decreasing sequence of elements, namely a copy of $Bbb{Zsetminus N}$, or a reverse copy of $Bbb N$. Let us denote by $omega^*$ this partial order.
Now consider $ntimesomega^*$, which is a consecutive chain of $n$ copies of $omega^*$. It has a maximal element, yes, but no minimal element.
What happens when you take $nneq m$ and consider $2^c$ disjoint copies of $ntimesomega^*$ and $2^c$ copies of $mtimesomega^*$? Are they isomorphic?
answered Nov 22 at 16:55
Asaf Karagila♦
301k32422753
Could you please explain with some more details? I'm not really sure if I understand what you're saying.
– mathbbandstuff
Nov 22 at 16:59
1
How about you tell me what you think I mean, and then I can clarify on the necessary points.
– Asaf Karagila♦
Nov 22 at 17:00
Of course. I understand how you defined $omega^*$. Is a partial order induced by the order of numbers in $mathbb{N}$? If yes, I understand that $ntimesomega^*$ has a maximal element, $(n, omega^*)$, if I'm correct. What I don't get is how do you mean to consider $2^c$ disjoint coipes of $ntimesomega^*$ and $mtimesomega^*$? How to treat more copies of $ntimesomega^*$? Do I need to compare them in the given order?
– mathbbandstuff
Nov 22 at 17:08
1
Taking $I$ disjoint copies of some order $P$ means that we consider $Itimes P$ with the order defined as $(i,p)leq (j,q)$ if and only if $i=j$ and $pleq_P q$. Here $I$ is some set of size $2^c$ and $P$ is $ntimesomega^*$ for some $n$.
– Asaf Karagila♦
Nov 22 at 17:14
1
@MakeTheTrumpetsBlow: Yes, it is the lexicographic product of ${0,ldots,n-1}$ with its usual ordering and $omega^*$.
– Asaf Karagila♦
Nov 22 at 17:42
|
show 6 more comments
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1 Answer
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active
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1 Answer
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active
oldest
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active
oldest
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active
oldest
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up vote
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accepted
Consider a decreasing sequence of elements, namely a copy of $Bbb{Zsetminus N}$, or a reverse copy of $Bbb N$. Let us denote by $omega^*$ this partial order.
Now consider $ntimesomega^*$, which is a consecutive chain of $n$ copies of $omega^*$. It has a maximal element, yes, but no minimal element.
What happens when you take $nneq m$ and consider $2^c$ disjoint copies of $ntimesomega^*$ and $2^c$ copies of $mtimesomega^*$? Are they isomorphic?
answered Nov 22 at 16:55
Asaf Karagila♦
301k32422753
Could you please explain with some more details? I'm not really sure if I understand what you're saying.
– mathbbandstuff
Nov 22 at 16:59
1
How about you tell me what you think I mean, and then I can clarify on the necessary points.
– Asaf Karagila♦
Nov 22 at 17:00
Of course. I understand how you defined $omega^*$. Is a partial order induced by the order of numbers in $mathbb{N}$? If yes, I understand that $ntimesomega^*$ has a maximal element, $(n, omega^*)$, if I'm correct. What I don't get is how do you mean to consider $2^c$ disjoint coipes of $ntimesomega^*$ and $mtimesomega^*$? How to treat more copies of $ntimesomega^*$? Do I need to compare them in the given order?
– mathbbandstuff
Nov 22 at 17:08
1
Taking $I$ disjoint copies of some order $P$ means that we consider $Itimes P$ with the order defined as $(i,p)leq (j,q)$ if and only if $i=j$ and $pleq_P q$. Here $I$ is some set of size $2^c$ and $P$ is $ntimesomega^*$ for some $n$.
– Asaf Karagila♦
Nov 22 at 17:14
1
@MakeTheTrumpetsBlow: Yes, it is the lexicographic product of ${0,ldots,n-1}$ with its usual ordering and $omega^*$.
– Asaf Karagila♦
Nov 22 at 17:42
|
show 6 more comments
up vote
3
down vote
accepted
Consider a decreasing sequence of elements, namely a copy of $Bbb{Zsetminus N}$, or a reverse copy of $Bbb N$. Let us denote by $omega^*$ this partial order.
Now consider $ntimesomega^*$, which is a consecutive chain of $n$ copies of $omega^*$. It has a maximal element, yes, but no minimal element.
What happens when you take $nneq m$ and consider $2^c$ disjoint copies of $ntimesomega^*$ and $2^c$ copies of $mtimesomega^*$? Are they isomorphic?
answered Nov 22 at 16:55
Asaf Karagila♦
301k32422753
Could you please explain with some more details? I'm not really sure if I understand what you're saying.
– mathbbandstuff
Nov 22 at 16:59
1
How about you tell me what you think I mean, and then I can clarify on the necessary points.
– Asaf Karagila♦
Nov 22 at 17:00
Of course. I understand how you defined $omega^*$. Is a partial order induced by the order of numbers in $mathbb{N}$? If yes, I understand that $ntimesomega^*$ has a maximal element, $(n, omega^*)$, if I'm correct. What I don't get is how do you mean to consider $2^c$ disjoint coipes of $ntimesomega^*$ and $mtimesomega^*$? How to treat more copies of $ntimesomega^*$? Do I need to compare them in the given order?
– mathbbandstuff
Nov 22 at 17:08
1
Taking $I$ disjoint copies of some order $P$ means that we consider $Itimes P$ with the order defined as $(i,p)leq (j,q)$ if and only if $i=j$ and $pleq_P q$. Here $I$ is some set of size $2^c$ and $P$ is $ntimesomega^*$ for some $n$.
– Asaf Karagila♦
Nov 22 at 17:14
1
@MakeTheTrumpetsBlow: Yes, it is the lexicographic product of ${0,ldots,n-1}$ with its usual ordering and $omega^*$.
– Asaf Karagila♦
Nov 22 at 17:42
|
show 6 more comments
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Consider a decreasing sequence of elements, namely a copy of $Bbb{Zsetminus N}$, or a reverse copy of $Bbb N$. Let us denote by $omega^*$ this partial order.
Now consider $ntimesomega^*$, which is a consecutive chain of $n$ copies of $omega^*$. It has a maximal element, yes, but no minimal element.
What happens when you take $nneq m$ and consider $2^c$ disjoint copies of $ntimesomega^*$ and $2^c$ copies of $mtimesomega^*$? Are they isomorphic?
answered Nov 22 at 16:55
Asaf Karagila♦
301k32422753
Consider a decreasing sequence of elements, namely a copy of $Bbb{Zsetminus N}$, or a reverse copy of $Bbb N$. Let us denote by $omega^*$ this partial order.
Now consider $ntimesomega^*$, which is a consecutive chain of $n$ copies of $omega^*$. It has a maximal element, yes, but no minimal element.
What happens when you take $nneq m$ and consider $2^c$ disjoint copies of $ntimesomega^*$ and $2^c$ copies of $mtimesomega^*$? Are they isomorphic?
answered Nov 22 at 16:55
Asaf Karagila♦
301k32422753
answered Nov 22 at 16:55
Asaf Karagila♦
301k32422753
answered Nov 22 at 16:55
Asaf Karagila♦
301k32422753
answered Nov 22 at 16:55
Asaf Karagila♦
301k32422753
301k32422753
Could you please explain with some more details? I'm not really sure if I understand what you're saying.
– mathbbandstuff
Nov 22 at 16:59
1
How about you tell me what you think I mean, and then I can clarify on the necessary points.
– Asaf Karagila♦
Nov 22 at 17:00
Of course. I understand how you defined $omega^*$. Is a partial order induced by the order of numbers in $mathbb{N}$? If yes, I understand that $ntimesomega^*$ has a maximal element, $(n, omega^*)$, if I'm correct. What I don't get is how do you mean to consider $2^c$ disjoint coipes of $ntimesomega^*$ and $mtimesomega^*$? How to treat more copies of $ntimesomega^*$? Do I need to compare them in the given order?
– mathbbandstuff
Nov 22 at 17:08
1
Taking $I$ disjoint copies of some order $P$ means that we consider $Itimes P$ with the order defined as $(i,p)leq (j,q)$ if and only if $i=j$ and $pleq_P q$. Here $I$ is some set of size $2^c$ and $P$ is $ntimesomega^*$ for some $n$.
– Asaf Karagila♦
Nov 22 at 17:14
1
@MakeTheTrumpetsBlow: Yes, it is the lexicographic product of ${0,ldots,n-1}$ with its usual ordering and $omega^*$.
– Asaf Karagila♦
Nov 22 at 17:42
|
show 6 more comments
Could you please explain with some more details? I'm not really sure if I understand what you're saying.
– mathbbandstuff
Nov 22 at 16:59
1
How about you tell me what you think I mean, and then I can clarify on the necessary points.
– Asaf Karagila♦
Nov 22 at 17:00
Of course. I understand how you defined $omega^*$. Is a partial order induced by the order of numbers in $mathbb{N}$? If yes, I understand that $ntimesomega^*$ has a maximal element, $(n, omega^*)$, if I'm correct. What I don't get is how do you mean to consider $2^c$ disjoint coipes of $ntimesomega^*$ and $mtimesomega^*$? How to treat more copies of $ntimesomega^*$? Do I need to compare them in the given order?
– mathbbandstuff
Nov 22 at 17:08
1
Taking $I$ disjoint copies of some order $P$ means that we consider $Itimes P$ with the order defined as $(i,p)leq (j,q)$ if and only if $i=j$ and $pleq_P q$. Here $I$ is some set of size $2^c$ and $P$ is $ntimesomega^*$ for some $n$.
– Asaf Karagila♦
Nov 22 at 17:14
1
@MakeTheTrumpetsBlow: Yes, it is the lexicographic product of ${0,ldots,n-1}$ with its usual ordering and $omega^*$.
– Asaf Karagila♦
Nov 22 at 17:42
Could you please explain with some more details? I'm not really sure if I understand what you're saying.
– mathbbandstuff
Nov 22 at 16:59
Could you please explain with some more details? I'm not really sure if I understand what you're saying.
– mathbbandstuff
Nov 22 at 16:59
1
1
How about you tell me what you think I mean, and then I can clarify on the necessary points.
– Asaf Karagila♦
Nov 22 at 17:00
How about you tell me what you think I mean, and then I can clarify on the necessary points.
– Asaf Karagila♦
Nov 22 at 17:00
Of course. I understand how you defined $omega^*$. Is a partial order induced by the order of numbers in $mathbb{N}$? If yes, I understand that $ntimesomega^*$ has a maximal element, $(n, omega^*)$, if I'm correct. What I don't get is how do you mean to consider $2^c$ disjoint coipes of $ntimesomega^*$ and $mtimesomega^*$? How to treat more copies of $ntimesomega^*$? Do I need to compare them in the given order?
– mathbbandstuff
Nov 22 at 17:08
Of course. I understand how you defined $omega^*$. Is a partial order induced by the order of numbers in $mathbb{N}$? If yes, I understand that $ntimesomega^*$ has a maximal element, $(n, omega^*)$, if I'm correct. What I don't get is how do you mean to consider $2^c$ disjoint coipes of $ntimesomega^*$ and $mtimesomega^*$? How to treat more copies of $ntimesomega^*$? Do I need to compare them in the given order?
– mathbbandstuff
Nov 22 at 17:08
1
1
Taking $I$ disjoint copies of some order $P$ means that we consider $Itimes P$ with the order defined as $(i,p)leq (j,q)$ if and only if $i=j$ and $pleq_P q$. Here $I$ is some set of size $2^c$ and $P$ is $ntimesomega^*$ for some $n$.
– Asaf Karagila♦
Nov 22 at 17:14
Taking $I$ disjoint copies of some order $P$ means that we consider $Itimes P$ with the order defined as $(i,p)leq (j,q)$ if and only if $i=j$ and $pleq_P q$. Here $I$ is some set of size $2^c$ and $P$ is $ntimesomega^*$ for some $n$.
– Asaf Karagila♦
Nov 22 at 17:14
1
1
@MakeTheTrumpetsBlow: Yes, it is the lexicographic product of ${0,ldots,n-1}$ with its usual ordering and $omega^*$.
– Asaf Karagila♦
Nov 22 at 17:42
@MakeTheTrumpetsBlow: Yes, it is the lexicographic product of ${0,ldots,n-1}$ with its usual ordering and $omega^*$.
– Asaf Karagila♦
Nov 22 at 17:42
|
show 6 more comments
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For an easier problem, how about if you instead wanted there to be $0$ maximal elements and $0$ minimal elements?
– Eric Wofsey
Nov 22 at 16:53