Existence of a countable family of pairwise non-similar partial orders with $2^c$ maximal elements, and no...











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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







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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

















up vote
1
down vote

favorite
1












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







share|cite|improve this question
























  • 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















up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1





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







share|cite|improve this question















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






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share|cite|improve this question













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edited Nov 22 at 17:08

























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




















  • 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












1 Answer
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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?






share|cite|improve this answer





















  • 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











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1 Answer
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1 Answer
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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?






share|cite|improve this answer





















  • 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















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?






share|cite|improve this answer





















  • 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













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?






share|cite|improve this answer












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?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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


















  • 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


















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