Prove there is either a chain or an antichain of infinite cardinal.

Question

Let $K$ be an set of infinite cardinal, $X$. Let $(K,\le)$ be a partially ordered set. Prove there is either a chain $C$ such that $|C|=X$ or there is an antichain $A$ such that $|A|=X$. I guess I should be using Zorn's lemma, but I don't know how. I would truly appreciate it if you helped me.

Answer 1

This is not true in general. For $n\in\omega$ let $\langle C_n,\le_n\rangle$ be a well-order of type $\omega_n$. Let $K$ be the disjoint union of the sets $C_n$, and let $\le$ be the union of the orders $\le_n$; $\le$ is a partial order on $K$. $K$ has cardinality $\omega_\omega$, but every chain in $K$ has cardinality $\omega_n<\omega_\omega$ for some $n\in\omega$, and every antichain in $K$ has cardinality at most $\omega$ (since it can contain at most one member of each $C_n$).

Answer 2

This is consistently false even for regular cardinals, a Suslin tree is a tree is size $\aleph_1$ with no uncountable chain or antichain. The existence of a Suslin tree is independent of the axioms of set theory.