Find a poset with no infinite antichains and that can only be covered by infinitely many chains.

Question

So far I'm struggling to find such a poset.

I first considered $(\mathbb{Z}, \leq )$ because it doesn't have any antichains so there aren't any infinite ones.

The problem is that whilst the chain it can be covered by is infinite, it only requires one chain rather than an infinite amount of chains.

Does anyone have any hints as to where I begin looking for such a poset?

Answer 1

The poset $\mathbb{N} \times \mathbb{N}$ has the desired property.

To show that it has no infinite antichains, use:

Definition 0. A pqo is a partially ordered set that is well-founded and has no infinite antichains.

Proposition 0. A finite product of pqo's is a pqo.

Following Nate's comment, we can show that it cannot be covered by finitely many chains using:

Definition 1/Proposition. Let $P$ denote a poset. Let $A$ denote an antichain of $P$ and $K$ denote a covering of $P$ by chains. Then we will write $[A,K]$ for the surjective partial function $A \leftarrow K$ that assigns to each $k \in K$ the unique $a \in A$ such that $a \in k$, whenever such an $a$ exists. Hence (proposition:) $|A| \leq |K|$.

Corollary. Let $P$ denote a poset that has an antichain of every finite cardinality. Then for every covering $K$ of $P$ by chains, we have $\aleph_0 \leq |K|.$