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We say that a partially ordered set $(P,\leq)$ is interval-isomorphic if for all $a<b \in P$ we have $P \cong [a,b]$, where $[a,b]=\{x\in P:a\leq x\leq b\}$.

Suppose $(P,\leq)$ is interval-isomorphic and there are $a,b\in P$ with $a<b$. Does this imply that $(P,\leq)$ is a lattice?

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    $\begingroup$ Is the $L$ in the definition of the interval $[a,b]$ supposed to be a $P$? $\endgroup$
    – Philipp Lampe
    Commented Sep 4, 2019 at 8:05
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    $\begingroup$ Right, thanks @PhilippLampe, I have just corrected this $\endgroup$
    – Dominic van der Zypen
    Commented Sep 4, 2019 at 8:11

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No. Let $P^-=\mathbb Q\times\{0,1\}$ with partial order defined by $$\langle x,a\rangle\le\langle y,b\rangle\iff x<y\lor(x=y\land a=b),$$ and let $P=P^-\cup\{-\infty,+\infty\}$ with $-\infty<\langle x,a\rangle<+\infty$.

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  • $\begingroup$ Very nice example, thanks! $\endgroup$
    – Dominic van der Zypen
    Commented Sep 4, 2019 at 9:07

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