# 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. – William Elliot May 12 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. – Mark Kamsma May 12 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 $f(\alpha)$? – Mark Kamsma May 12 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. – Trusio May 12 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? – Alex Kruckman May 12 at 22:06