Erratum to "On Operations and Linear Extensions of Well Partially Ordered Sets"

TL;DR

This paper provides a counter-example to a previously claimed lemma in the study of operations and linear extensions of well partially ordered sets, challenging prior assumptions in the field.

Contribution

It introduces a specific counter-example that refutes Lemma 12 from the original work, clarifying the limitations of the earlier results.

Findings

Counter-example disproves Lemma 12

Highlights need for revised assumptions in the theory

Refines understanding of operations on well partial orders

Abstract

In this article, we give a counter-example to Lemma 12 of the article "On Operations and Linear Extensions of Well Partially Ordered Sets" by Maciej Malicki and Aleksander Rutkowski.

Full text

Current version : 2019/04/28

1 Introduction

In this article, we give a counter-example to Lemma 12 of the article “On Operations and Linear Extensions of Well Partially Ordered Sets” by Maciej Malicki and Aleksander Rutkowski (Malicki and Rutkowski (2004)).

2 Definitions and notations

Definition 2.1** (Rank function).**

Each well-founded poset $P$ admits an ordinal valued rank function $\operatorname{rank}_{P}$ defined inductively on its elements: $\operatorname{rank}_{P}(a)=sup_{x<_{P}a}(\operatorname{rank}_{P}(x)+1)$

Let $\mathcal{P}=\{P_{t}:t\in T\}$ be an ordered family of ordered sets, i.e. both $P_{t}$’s and $T$ are partially ordered (by $\leq_{t}$ and $\leq_{T}$ respectively). With no loss of generality, elements of $\mathcal{P}$ can be assumed to be pairwise disjoint. Let, for $a \in\bigcup_{t\in T}P_{t}$, $f(a)$ be that unique $t$ such that $a\in P_{t}$.

Now, assume all elements of $\mathcal{P}$ to be well-founded and call, for $a \in\bigcup_{t\in T}P_{t}$, the primitive rank of $a$ an ordinal $g(a) =\operatorname{rank}_{P_{f(a)}}(a)$. Define the following ranked order $<_{R}$ on $\bigcup_{t\in T}P_{t}$: $a <_{R}b$ if

• •

either $f(a)=f(b)$ and $a <_{f(a)}b$,

• •

or $f(a)<f(b)$ and $g(a)\leq g(b)$.

Call the union with that order the ranked sum and denote it $\mathcal{RP}$. Observe that $a \leq_{R}b$ implies $f(a) \leq_{T}f(b)$.

3 A counter example to Lemma 12

False lemma 3.1** (Lemma 12).**

Let both $T$ and all the components $P_{t}$ of $\mathcal{RP}$ be well-founded (hence $P=\mathcal{RP}$ is well-founded too). Then for each $a \in\bigcup_{t\in T}P_{t}$, $\operatorname{rank}_{P}(a)\leq\operatorname{rank}_{T}(f(a))+\operatorname{rank}_{f(a)}(a)$.

Counter-example:

• It is easy to construct an order $\mathcal{RP}$ with an element $a$ such that $\operatorname{rank}_{P}(a)=\operatorname{rank}_{T}(f(a))+\operatorname{rank}_{f(a)}(a)+1$. Indeed consider $T=\{0,1\}$, and $P_{0}=P_{1}=\omega+1$ ($\omega$ is the first infinite ordinal). Let $a$ be the maximum of $P_{1}$, and $b$ be the maximum of $P_{0}$. Then $\operatorname{rank}_{\mathcal{RP}}(b)=\operatorname{rank}_{P_{0}}(b)$, hence $\operatorname{rank}_{\mathcal{RP}}(a)=\operatorname{rank}_{\mathcal{RP}}(b)+1=\omega+1>\operatorname{rank}_{T}(f(a))+\operatorname{rank}_{f(a)}(a)=\operatorname{rank}_{T}(P_{1})+\operatorname{rank}_{P_{1}}(a)=1+\omega=\omega$ (ordinal sum is not commutative and $1+\omega\neq\omega+1$).

The problem in the proof is in the line $\operatorname{rank}_{T}(f(b))+\operatorname{rank}_{f(b)}(b)+1\leq\operatorname{rank}_{T}(f(a))+\operatorname{rank}_{f(a)}(a)$. It should be corrected to $\operatorname{rank}_{T}(f(b))+1+\operatorname{rank}_{f(b)}(b)\leq\operatorname{rank}_{T}(f(a))+\operatorname{rank}_{f(a)}(a)$, but then the proof by transfinite induction fails.

You cannot correct the lemma by switching both ranks, i.e. $\operatorname{rank}_{P}(a)\leq\operatorname{rank}_{f(a)}(a)+\operatorname{rank}_{T}(f(a))$. Indeed then $T=\omega+1$, and $P_{0}=P_{1}=...=P_{\omega}=\{0,1\}$ is a counter-example.

Lemma 3.2**.**

For any ordinal $\alpha$, there is an order $\mathcal{RP}$ with an element $a$ such that $\operatorname{rank}_{P}(a)=\operatorname{rank}_{T}(f(a))+\operatorname{rank}_{f(a)}(a)+\alpha$.

Proof.

Consider $T=\alpha+1$, and $P_{0}=P_{1}=...=P_{\alpha}=\beta+1$, where $\beta$ is the first ordinal such that $\alpha+\beta=\beta$. Let $a$ be the maximum of $P_{\alpha}$, and $b$ be the maximum of $P_{0}$. Then $\operatorname{rank}_{\mathcal{RP}}(b)=\operatorname{rank}_{P_{0}}(b)$, hence $\operatorname{rank}_{\mathcal{RP}}(a)=\operatorname{rank}_{\mathcal{RP}}(b)+\alpha=\beta+\alpha>\operatorname{rank}_{T}(f(a))+\operatorname{rank}_{f(a)}(a)=\operatorname{rank}_{T}(P_{\alpha})+\operatorname{rank}_{P_{\alpha}}(a)=\alpha+\beta=\beta$.

Lemma 3.3**.**

For any ordinal $\alpha$, there is an order $\mathcal{RP}$ with an element $a$ such that $\operatorname{rank}_{P}(a)=\operatorname{rank}_{f(a)}(a)+\operatorname{rank}_{T}(f(a))+\alpha$.

4 Conclusion

We sent an email to one of the authors on 2019/02/24 but, unfortunately, we never had an answer. We hope this erratum may be useful to the scientific community.

Acknowledgements.

We thank God: Father, Son, and Holy Spirit. We thank Maria. They help us through our difficulties in life.