I know that there is an infinite partially ordered set that violates Dilworth's Theorem and one that violates Mirsky's Theorem. Unfortunately, I do not have access to the references for these counterexamples. I have a few questions.
Can somebody please provide an explicit counterexample for each theorem?
I would like to know if there is an infinite partially ordered set that violates both theorems. I am curious whether these theorems are true duals in the sense that a partially ordered set satisfies one of them if and only if it satisfies both.
Let me call a partially ordered set $P$ Dilworth if the width of $P$ (i.e., the supremum of the cardinalities of antichains in $P$) is the same as the smallest cardinal number $c$ such that there exists a family $\mathcal{C}$ of chains in $P$ such that $P=\bigcup\mathcal{C}$ and $|\mathcal{C}|=c$. A partially ordered set $P$ is Mirsky if the height of $P$ (i.e., the supremum of the cardinalities of chains in $P$) is the same as the smallest cardinal number $a$ such that there exists a family $\mathcal{A}$ of antichains in $P$ such that $P=\bigcup\mathcal{A}$ and $|\mathcal{A}|=a$.
For example, finite and countably infinite partially ordered sets are both Dilworth and Mirsky. More generally, a partially ordered set of finite width is Dilworth, whereas a partially ordered set of finite height is Mirsky. That is, the questions above can be rephrased as follows.
Please give me an example of a non-Dilworth partially ordered set and an example of a non-Mirsky partially ordered set.
Is there a partially ordered set which is simultaneously non-Dilworth and non-Mirsky? Does it hold that a partially ordered set is Dilworth if and only if it is Mirsky?
1a. Let $\kappa$ be an uncountable cardinal. Partially order the Cartesian product $P_\kappa=\kappa\times\kappa$ so that $$\langle\alpha,\beta\rangle\le\langle\alpha',\beta'\rangle\iff\alpha\le\alpha'\ \text{ and }\ \beta\le\beta'.$$ Then every antichain in $P_\kappa$ is finite, but $P_\kappa$ is not the union of $\lt\kappa$ chains. As $P_\kappa$ has cardinality $\kappa$ and height $\kappa,$ it is Mirsky but not Dilworth.
1b. Let $\kappa$ be an uncountable cardinal. Partially order the Cartesian product $Q_\kappa=\kappa\times\kappa$ so that $$\langle\alpha,\beta\rangle\lt\langle\alpha',\beta'\rangle\iff\alpha\lt\alpha'\ \text{ and } \beta\gt\beta'.$$ then every chain in $Q_\kappa$ is finite, but $Q_\kappa$ is not the union of $\lt\kappa$ antichains. As $Q_\kappa$ has cardinality $\kappa$ and width $\kappa,$ it is Dilworth but not Mirsky.
2a. If all chains and antichains in a partially ordered set $P$ are finite, then $P$ is finite by Ramsey's theorem. However, there is a poset $P$ of cardinality $2^{\aleph_0}$ in which all chains and antichains are countable; of course such a $P$ is not the union of countably many chains or antichains, so it is neither Dilworth not Mirsky. The example is due to Sierpiński: take the intersection of two linear orderings of $\mathbb R,$ the usual ordering and a well-ordering.
2b. I've already given natural examples of posets which are Mirsky but not Dilworth or Dilworth but not Mirsky. Still, it may be worth pointing out that, given a poset $P$ which is neither Dilworth nor Mirsky, we can trivially make it Mirsky-but-not-Dilworth by adjoining a large chain, or make it Dilworth-but-not-Mirsky by adjoining a large antichain.
Of course, any poset of finite height or width is both Dilworth and Mirsky.