TITLE: Prove that for any infinite poset there is an infinite subset which is either linearly ordered or antichain.
QUESTION [10 upvotes]: Let $(X, \leq)$ be an infinite poset. Prove that there is an infinite $X' \subset X$ for which holds one of the following:
1) Induced order on $X'$ is linear.
2) $(X', \leq_\mathrm{ind})$ is antichain. (Every two elements in $X'$ are incomparable)

My attempt:
Let's start trying to construct a linearly ordered set $(X', \leq_\mathrm{ind})$ element by element. First we take some element $x_1 \in X$, then we take another $x_2 \in X$ such as $x_1 \leq x_2$ or $x_2 \leq x_1$. Basically, we take an element from $X$, if it is "good" we put it in our chain, otherwise we throw it away. We continue constructing the chain that way. If we do not stop at some finite step, then we are done, we have constructed linearly ordered subset.
If we do stop at some $x_n$ that means that we have finite amount of comparable elements, which means that $X \setminus X'$ is an antichain.

a) I am not sure about the soundness in the last step, could somebody please check it?
b) Can this be proved without a choice function?

REPLY [7 votes]: One can use a free ultrafilter to get a reasonably slick argument that minimizes the amount of detail that has to be kept track of at the cost of using a different consequence of AC $-$ the Boolean prime ideal theorem, which is strictly weaker than AC and independent of Dependent Choice, which also suffices.
Let $\mathscr{U}$ be a free ultrafilter on $X$. For each $x\in X$ let $C(x)=\{y\in X:x\le y\lor y\le x\}$, and let $A(x)=X\setminus C(x)=\{y\in X:x\not\le y \land y\not\le x\}$; exactly one of $C(x)$ and $A(x)$ belongs to $\mathscr{U}$. Let $C=\{x\in X:C(x)\in\mathscr{U}\,\}$, and let $A=X\setminus C=\{x\in X:A(x)\in\mathscr{U}\,\}$. Exactly one of $A$ and $C$ belongs to $\mathscr{U}$; without loss of generality assume that $C\in\mathscr{U}$.
Fix $x_0\in C$, and let $C_0=C\cap C(x_0)\setminus\{x_0\}\in\mathscr{U}$. Given $C_n\in\mathscr{U}\,$ for some $n\in\omega$, choose $x_{n+1}\in C_n$ and let $C_{n+1}=C_n\cap C(x_{n+1})\setminus\{x_{n+1}\}\in\mathscr{U}\,$; clearly the construction goes through to $\omega$ to produce a set $\{x_n:n\in\omega\}$ of distinct elements of $X$. Moreover, for any $k<n<\omega$ we have $x_n\in C_k\subseteq C(x_k)$, so $\{x_n:n\in\omega\}$ is an infinite chain in $X$. (Had $A$ been a member of $\mathscr{U}\,$ instead of $C$, the set $\{x_n:n\in\omega\}$ would of course have been an infinite antichain.)
As JDH has already noted, this is essentially just the two-color case of the infinite Ramsey theorem, $\kappa\to(\omega)_2^2$. If one wanted to use a really big stick, one could beat the problem over the head with the Erdős-Dushnik-Miller theorem, $\kappa\to(\kappa,\omega)^2$, and conclude that either $X$ has a chain of cardinality $|X|$, or $X$ has an infinite antichain. (Here the rôles of chain and antichain can of course be reversed.)