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 $|A'|>|A|$?

Improve this question
$\endgroup$
3
  • $\begingroup$ 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
    Commented Oct 4, 2017 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
    Commented Oct 4, 2017 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
    Commented Oct 4, 2017 at 15:28

1 Answer 1

Reset to default
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.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .