3
$\begingroup$

Let $(S,<)$ be a countably infinite semilattice such that:

1) $S$ is no-where dense - i.e. there does not exist a subset $T$ such that for all $a,b\in T$ with $a<b$, there exists $c\in T$ with $a<c<b$,

2) $S$ has finite width $n$ - i.e. the maximum size of the anti-chains of $S$ is $n$.

Let $a,b\in S$ be such that $a<b$.

Question: Is the set $\{x\in S :a<x<b \}$ finite?

$\endgroup$
2
$\begingroup$

I think the simplest example is given by the ordinal $\omega+1=\{0<1<2<\cdots<\omega\}$. As a well order it contains no antichains, and it is well founded (hence nowhere dense), but of course $\{x\in\omega+1\mid 0<x<\omega\}$ is infinite. Ever infinite well order would do. You will find plenty of examples in the larger class of well quasi-orders too.

$\endgroup$

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.