Let $(A,\geq)$ be a partially ordered set such that

  • there exists the join $\bigvee A$, i.e. $a\in A$ such that $a\geq b$ for any $b\in A$;
  • for any pair $(b,c)\in A\times A$ there exists the meet $b \wedge c\in A$, i.e. $d\in A$ such that $d\leq b,c$ and such that $d\geq e$ for any other $e\in A$ satisfying $e\leq b,c$.

Is it possible to construct a bijection $f: \eta\to A$ for $\eta$ ordinal number such that for any $\alpha<\beta\leq \eta$ there exists $\gamma\leq\beta$ such that $$ f(\alpha)\wedge f(\beta)= f(\gamma)\, \,? $$ May you suggest some references for my problem? I am new in the field of set theory.

  • $\begingroup$ $ \eta=\big\{1,\dots, \{\eta\} \big\}$ violates regularity, foundations. $\endgroup$ May 12, 2019 at 12:05
  • $\begingroup$ There are a few things off about your notation. William already pointed one out. Also, ordinals start at $0$ (so it would be $\eta = \{0, \ldots\}$). Also, what are your restrictions on $\eta$? If you just want some ordinal, then a few trivially work (e.g. $0$ or $1$). If you want this constructions for all ordinals, then that is not possible. $\endgroup$ May 12, 2019 at 15:15
  • $\begingroup$ Oh, and if $f$ is a function $\eta \to A$, then $f(\{\alpha\})$ does not make sense in general. Do you mean $f(\alpha)$? $\endgroup$ May 12, 2019 at 15:16
  • $\begingroup$ Sorry, you were right. I edited the question. As I said I am new in set theory. Anyway Mark, the restriction on $\eta$ is that it has to be in bijection with the poset A. $\endgroup$
    – Trusio
    May 12, 2019 at 21:48
  • $\begingroup$ I wonder whether you're really asking the question you want to be asking. Can you give us some more context about why you want to know the answer? $\endgroup$ May 12, 2019 at 22:06

1 Answer 1


No, this is not true in general. Let $(A,\leq)$ be a poset with the two required properties such that every maximalan infinite decreasing $\leq$-chain is not well-founded, for example if each maximal $\leq$-chain is isomorphic to $\{-\infty\}\cup(\mathbb Z\setminus\omega)$. (I believe this shouldn't be too hard to construct.) Now that we have this "very non-well-founded" poset, assume towards a contradiction that such a bijection $f$ exists. Since $f$ is a bijection, when we have $f(\alpha)\wedge f(\beta)=f(\gamma)$, if $f(\gamma)\lneq f(\beta)$ then we have $\gamma\lneq\beta$. So the image of any maximal $\leq$-chain must have an infinite decreasing sequence under $\lneq$, but this is impossible since the image is a set of ordinals.

In order to make a weaker variant of this question that is true, requiring some sort of well-quasi-ordering condition for the poset might be relevant, to avoid the image of $f$ having an infinite decreasing sequence of ordinals like this.


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