1
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ Is the $L$ in the definition of the interval $[a,b]$ supposed to be a $P$? $\endgroup$ – Philipp Lampe Sep 4 '19 at 8:05
  • 1
    $\begingroup$ Right, thanks @PhilippLampe, I have just corrected this $\endgroup$ – Dominic van der Zypen Sep 4 '19 at 8:11
4
$\begingroup$

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 $-\infty<\langle x,a\rangle<+\infty$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.