# Are non-trivial interval-isomorphic posets lattices?

We say that a partially ordered set $$(P,\leq)$$ is interval-isomorphic if for all $$a we have $$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 $$a. Does this imply that $$(P,\leq)$$ is a lattice?

• Is the $L$ in the definition of the interval $[a,b]$ supposed to be a $P$? – Philipp Lampe Sep 4 '19 at 8:05
• Right, thanks @PhilippLampe, I have just corrected this – Dominic van der Zypen Sep 4 '19 at 8:11

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