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

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.

• For an easier problem, how about if you instead wanted there to be $0$ maximal elements and $0$ minimal elements? – Eric Wofsey Nov 22 '18 at 16:53

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?

• Could you please explain with some more details? I'm not really sure if I understand what you're saying. – mathbbandstuff Nov 22 '18 at 16:59
• How about you tell me what you think I mean, and then I can clarify on the necessary points. – Asaf Karagila Nov 22 '18 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 $n\times\omega^*$ 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? – mathbbandstuff Nov 22 '18 at 17:08
• 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 $n$. – Asaf Karagila Nov 22 '18 at 17:14
• @MakeTheTrumpetsBlow: Yes, it is the lexicographic product of $\{0,\ldots,n-1\}$ with its usual ordering and $\omega^*$. – Asaf Karagila Nov 22 '18 at 17:42