3
$\begingroup$

The following construction gives a poset such that no antichain has maximum cardinality: For $n\in\mathbb{N}\setminus\{0\}$, let "layer" $n$ consist of an antichain of $n$ points, and as for the ordering: if $m<n \in \mathbb{N}\setminus\{0\}$ every point of layer $m$ is smaller than every point in layer $n$.

This poset is far from being a lattice. Which leads to the

Question. Is there a lattice $(L,\leq)$ such that for every antichain $A\subseteq L$ there is an antichain $A' \subseteq L$ with $|A'|>|A|$?

$\endgroup$
  • $\begingroup$ Lattices do not have to have a top element. Consider the 3 smooth numbers under divisibility. Gerhard "Likes General Algebraic Number Theories" Paseman, 2017.10.04. $\endgroup$ – Gerhard Paseman Oct 4 '17 at 15:24
  • $\begingroup$ Right - I will delete my remark about the top element. What are smooth numbers? Do they give an example for a lattice as in the question? Thanks! $\endgroup$ – Dominic van der Zypen Oct 4 '17 at 15:25
  • $\begingroup$ 3-smooth numbers are another name for positive integers of the form $2^a3^b$, for integral exponents $a$ and $b$. Divisibility in the natural numbers is a lattice order. Gerhard "Perhaps I Should Say 'Friable'" Paseman, 2017.10.04. $\endgroup$ – Gerhard Paseman Oct 4 '17 at 15:28
6
$\begingroup$

Yes. For $n \in \mathbb{N}$ even let "layer" $n$ consist of a single point, and for $n \in \mathbb{N}$ odd let "layer" $n$ consist of $n$ points. The ordering is the same as yours, one point is less than another if and only if it lies in a lower layer.

$\endgroup$

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.