Is there a poset with 0 with countable automorphism group? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T07:16:04Z http://mathoverflow.net/feeds/question/7951 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/7951/is-there-a-poset-with-0-with-countable-automorphism-group Is there a poset with 0 with countable automorphism group? Harrison Brown 2009-12-06T03:03:46Z 2009-12-06T03:50:51Z <p>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?</p> <p>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.</p> http://mathoverflow.net/questions/7951/is-there-a-poset-with-0-with-countable-automorphism-group/7954#7954 Answer by David Speyer for Is there a poset with 0 with countable automorphism group? David Speyer 2009-12-06T03:34:34Z 2009-12-06T03:34:34Z <p>What about $\mathbb{Z} \cup \{ - \infty \}$? Here $- \infty$ is less than everything, and $\mathbb{Z}$ has the usual order.</p> <p>You probably want some sort of descending chain condition, to rule this out.</p> http://mathoverflow.net/questions/7951/is-there-a-poset-with-0-with-countable-automorphism-group/7955#7955 Answer by Hugh Thomas for Is there a poset with 0 with countable automorphism group? Hugh Thomas 2009-12-06T03:50:51Z 2009-12-06T03:50:51Z <p>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 $0$. </p> <p>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. </p> <p>By d.c.c., every element of $P$ is in some $P_n$. </p> <p>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 $H_n$. </p> <p>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. </p>