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.
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.