Been thinking about this one for awhile and I'm still stuck... Thought about proving it by arriving at a contradiction but I haven't reached anything noteworthy. If you do a proof by contradiction, you're assuming that you have an infinite poset that does not contain an infinite chain and infinite totally unordered set but I don't know how that helps me :(
2026-03-26 18:50:45.1774551045
Given an infinite poset, show that it contains either a infinite chain or an infinite totally unordered set.
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I’m assuming that by totally unordered set you mean an antichain.
Let $\langle P,\preceq\rangle$ be your poset, and let $P_0$ be a countably infinite subset of $P$. $[P_0]^2$ is the set of two-element subsets of $P_0$. Define a map $c:[P_0]^2\to\{0,1\}$ by setting $c(\{x,y\})=1$ iff $x$ and $y$ are comparable (i.e., $p\preceq q$ or $q\preceq p$), and apply the infinite Ramsey theorem.