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Poset Exponentiation and a Counterexample Birkhoff Said in 1942 He Did Not Have

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Abstract

Let A, P, and Q ≠ ∅ be posets and consider Professor Garrett Birkhoff’s exponentiation operator PQ. McKenzie defined a new operator

$$ \mathcal C\left( P^{Q}\right):=\{f\in P^{Q}\mid f\ \text{is in the same connected component as some}\ g $$
$$ \text{that is constant on the connected components of}\ Q\}. $$

He implied without proof that \(\mathcal C\left (A^{P\times Q}\right )\cong \mathcal C\left (\mathcal C\left (A^{P}\right )^{Q}\right )\), stating explicitly that this isomorphism was via the canonical order-isomorphism \({\Psi }:A^{P\times Q}\cong \left (A^{P}\right )^{Q}\) when A was connected. In this note, it is shown that the restriction of Ψ does not in general give an order-isomorphism from \(\mathcal C\left (A^{P\times Q}\right )\) to \(\mathcal C\left (\mathcal C\left (A^{P}\right )^{Q}\right )\), not even for A connected, although it does if AP has finite diameter or if Q has finitely many components. In his 1942 article on the arithmetic of ordered sets, Professor Birkhoff proved that if P embeds in Q and A is a complete lattice, then AP embeds in AQ. He said, “The author has no counterexample to [this theorem] for [posets] not complete lattices….” In this note, counterexamples where A, P, and Q are lattices are found.

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Acknowledgements

The author would like to thank Dr. Dominic van der Zypen of the Swiss Armed Forces, and the author thanks the referee for his or her remarks.

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Correspondence to Jonathan David Farley.

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Farley, J.D. Poset Exponentiation and a Counterexample Birkhoff Said in 1942 He Did Not Have. Order 39, 243–250 (2022). https://doi.org/10.1007/s11083-021-09564-5

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