# Given a poset with a join such that any pair has a meet, find a total order with a special property.

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.

• $\eta=\big\{1,\dots, \{\eta\} \big\}$ violates regularity, foundations. May 12, 2019 at 12:05
• 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. May 12, 2019 at 15:15
• Oh, and if $f$ is a function $\eta \to A$, then $f(\{\alpha\})$ does not make sense in general. Do you mean $f(\alpha)$? May 12, 2019 at 15:16
• 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. May 12, 2019 at 21:48
• 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? May 12, 2019 at 22:06

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.