3
$\begingroup$

Suppose $(P,\leq)$ is a poset without maximal elements. For $X\subseteq P$ we set $X^u = \{p\in P: p \geq x \text{ for all } x\in X\}$ and call this the set of upper bounds of $X$. We say that $B\subseteq P$ is unbounded if $B^u = \emptyset$. Moreover we say $D\subseteq P$ is dominating if for all $p\in P$ there is $d\in D$ such that $p\leq d$. We set

  • ${\frak b}(P) = \min\{|B|: B\subseteq P\text{ is unbounded}\}$, and
  • ${\frak d}(P) = \min\{|D|: B\subseteq P\text{ is dominating}\}$.

It is easy to see that for all posets $P$ without maximal elements we have ${\frak b}(P) \leq {\frak d}(P)$. What is an example of a poset $P$ in which we can prove in $\textsf{ZFC}$ that ${\frak b}(P) < {\frak d}(P)$?

$\endgroup$
7
$\begingroup$

$\omega\times \omega_1$ with the product (pointwise) order.

$\endgroup$
  • $\begingroup$ O frabjous day! Callooh! Callay! (The 6th upvote for this answer lifted me above the 10k mark. Not my deepest answer, though.) $\endgroup$ – Goldstern Oct 17 '17 at 18:05
  • $\begingroup$ (But a very concise one - and congrats!) $\endgroup$ – Dominic van der Zypen Nov 5 '17 at 17:42

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.