# Is there a poset with 0 with countable automorphism group?

Is there a poset P with a unique least element, such that every element is covered by finitely many other elements of P (and P is locally finite -- actually, per David Speyer's example, let's say that it satisfies the descending chain condition), and P has countably infinite automorphism group?

The question is motivated by extensions of Sperner's theorem and the LYM inequality to infinite posets. In particular I'm interested in whether you can extend Bollobas' (I believe) probabilistic proof of LYM to the infinite setting in general -- you can for some specific posets. But a prerequisite for a direct extension is for the automorphism group of the poset to be compact, and at the very least we want it to have some nice topological properties. So a poset with countably infinite automorphism group would be a very interesting case.

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It seems unlikely (once you assume d.c.c.). Define the height of an element $x$ in $P$ to be the length of the shortest unrefinable chain from $x$ to $0$.

Let $P_n$ denote the elements of $P$ whose height is at most $n$. Since each element has a finite number of covers, the number of elements in $P_n$ is finite.

By d.c.c., every element of $P$ is in some $P_n$.

Let $G$ denote the automorphisms of $P$ and let $G_n$ denote the automorphisms of $P_n$. $G$ is the inverse limit of the system $G_n$. Let $H_n$ denote the image of $G$ inside $G_n$. (Note that this might not be all of $G_n$, since there could be automorphisms of $P_n$ that don't extend to $P$.) $G$ is also the inverse limit of the system $H_n$.

If the system $H_n$ stabilizes, then $G$ is finite. On the other hand, if $H_n$ doesn't stabilize, then the cardinality of $G$ is an infinite product, i.e. uncountable.

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Cool! This is more or less what my intuition was, although I didn't know the bit of abstract nonsense about inverse systems that lets you formalize it. Just to be clear, though -- by "it seems unlikely" you mean "no, and here's why not", right? :) –  Harrison Brown Dec 6 '09 at 4:00
Yeah. I got more definite as I went along. –  Hugh Thomas Dec 6 '09 at 4:19
What about $\mathbb{Z} \cup \{ - \infty \}$? Here $- \infty$ is less than everything, and $\mathbb{Z}$ has the usual order.