How many non-isomorphic partial orders of a certain size?

I'm working on the consistency of Martin's axiom and I need some help counting. Assume we are in a universe where GCH holds and $\kappa$ is a regular cardinal. How many non-isomorphic partial orders are there of size less than $\kappa$. I think the answer is $\kappa$ (indeed this is what Jech writes) and can convince myself, but it is not very clear to me how to systematically count the number of different partial orders of a certain size.

Let me know if I need to give more context, I'm asking my specific question in a general way.

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hmm I think I figured it out. I beleive the way to do it is by counting possible chains (and thinking of them as subsets). I'd still like an expert to explain it clearly. – Erin K Carmody Jan 31 '12 at 22:37
I'm far from an expert, but I hope my answer explains it clearly. – Asaf Karagila Jan 31 '12 at 22:47

Suppose $(P,\le)$ is a partial order and $|P|<\kappa$. We can embed every such $P$ in its power set, using $p\in P$ then $p\mapsto\{x\in P\mid x\le p\}$. If we assume GCH then without loss of generality $P=\lambda<\kappa$ then $P(\lambda)=\lambda^+\le\kappa$.

Now we have that every partial order of cardinality less than $\kappa$ can be embedded into its power set, which is at most $\kappa$.

Note that every possible partially ordered set can be embedded this way into its power set, thus:

1. If $\kappa$ is a limit cardinal (and of course regular) every power set is smaller, the number of partially ordered sets is at most $\lambda^+$ for $\lambda<\kappa$, and thus we have $\kappa^{<\kappa}=\kappa$ many partially ordered sets of size $<\kappa$.
2. If $\kappa=\mu^+$ is a successor cardinal (thus regular assuming choice) then every partially ordered set can be embedded in $P(\mu)=\kappa$ - by GCH - and so there are only $\kappa$ many of those.
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yes thank you. I have a follow-up question. I'll ask soon. – Erin K Carmody Jan 31 '12 at 23:24
I am just heading to bed, so I'll only see that much later. I hope it's fine. – Asaf Karagila Jan 31 '12 at 23:26
I believe you have an upper bound. A lower bound is the number of linear orders which is the same as the upper bound. – Erin K Carmody Feb 5 '12 at 4:38
@Erin: Yeah, in the case of successor cardinal you can think of it as an upper bound; the lower bound is trivially achieved. Fix one order type for $\mu$ and then every permutation of $\mu$ defines a new partially ordered set; there are $\mu^+=\kappa$ many of those. – Asaf Karagila Feb 5 '12 at 6:27