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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.

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  • $\begingroup$ For an easier problem, how about if you instead wanted there to be $0$ maximal elements and $0$ minimal elements? $\endgroup$ – Eric Wofsey Nov 22 '18 at 16:53
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Consider a decreasing sequence of elements, namely a copy of $\Bbb{Z\setminus N}$, or a reverse copy of $\Bbb N$. Let us denote by $\omega^*$ this partial order.

Now consider $n\times\omega^*$, 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 $n\neq m$ and consider $2^c$ disjoint copies of $n\times\omega^*$ and $2^c$ copies of $m\times\omega^*$? Are they isomorphic?

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  • $\begingroup$ Could you please explain with some more details? I'm not really sure if I understand what you're saying. $\endgroup$ – mathbbandstuff Nov 22 '18 at 16:59
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    $\begingroup$ How about you tell me what you think I mean, and then I can clarify on the necessary points. $\endgroup$ – Asaf Karagila Nov 22 '18 at 17:00
  • $\begingroup$ 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 $n\times\omega^*$ 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 $n\times\omega^*$ and $m\times\omega^*$? How to treat more copies of $n\times\omega^*$? Do I need to compare them in the given order? $\endgroup$ – mathbbandstuff Nov 22 '18 at 17:08
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    $\begingroup$ Taking $I$ disjoint copies of some order $P$ means that we consider $I\times P$ with the order defined as $(i,p)\leq (j,q)$ if and only if $i=j$ and $p\leq_P q$. Here $I$ is some set of size $2^c$ and $P$ is $n\times\omega^*$ for some $n$. $\endgroup$ – Asaf Karagila Nov 22 '18 at 17:14
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    $\begingroup$ @MakeTheTrumpetsBlow: Yes, it is the lexicographic product of $\{0,\ldots,n-1\}$ with its usual ordering and $\omega^*$. $\endgroup$ – Asaf Karagila Nov 22 '18 at 17:42

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