### Three counterexamples concerning ω-chain completeness and fixed point properties

J. D. Mashburn
1981 Proceedings of the Edinburgh Mathematical Society
A partially ordered set, is -chain, in P, the least upper bound of C, denoted by sup C, exists. Notice that C could be empty, so an o> -chain complete partially ordered set has a least element, denoted by 0. A function / mapping a partially ordered set P into a partially ordered set Q is chain continuous if for any nonempty chain C in P, which has a supremum, /(sup P C) = sup Q /(C). It is o>-chain continuous if for any nonempty countable chain C in P, which has a supremum, /(sup P C) = sup Q
more » ... (sup P C) = sup Q /(C). A partially ordered set P has the least fixed point property if every order-preserving function from P to itself has a least fixed point. It has the fixed point property if every order-preserving function from P to itself has a fixed point. It has the least fixed point property for o>-chain continuous functions if every -chain completeness follows from the least fixed point property for a) -chain continuous functions. Kolodner has shown in (5) that the converse is true. The following example shows that the answer to Plotkin's question is no. Example 1. Let P l be the partially ordered set defined by the diagram on the next page. Let C = {c n |neN}, X = { x J n e N } and Y = {y n |neN}. Then x = s u p X and C has no supremum in P. Proposition 1. Every <o-chain continuous function from P x to itself has a least fixed point.