Let $(S,<)$ be a partially ordered set such that $S$ is uncountable and has the property that for all $x\in S$ there exists a $y\in S$ such that $y<x$. Is it always true that $S$ contains an uncountable chain (with or without assuming the axiom of choice)?
My basic approach would be to start with any $x_0\in S$ then pick a $x_1$ guaranteed by the condition such that $x_1<x_0$ and repeat the process indefinitely. However, I don't know if this is valid because this process is sequential and uncountable sets are not. I know that in assuming the axiom of choice, there would also be a well ordering on $S$ but this might not help at all because the poset may not extend to one. Any suggestions or counterexamples?
Best Answer
No, this isn't true: consider uncountably many copies of (say) $\mathbb{Z}$ with the usual ordering, placed "side-by-side."
(Concretely: let $I$ be some uncountable set, and let $\mathbb{P}$ be the partial order with underlying set $I\times\mathbb{Z}$ ordered by $$(i,a)\trianglelefteq(j,b)\quad\iff\quad i=j\mbox{ and }a\le b$$ where $\le$ is the usual ordering on $\mathbb{Z}$.)
That said, here are a couple quick comments (in $\mathsf{ZFC}$):
If we strengthen the requirement to "For every countable chain $X\subseteq S$ there is a $y\in S$ with $y\trianglelefteq x$ for each $x\in X$," then the answer is affirmative: by transfinite recursion we can build a descending sequence of ordertype $\omega_1$.
Meanwhile, that the counterexample above, while lacking long chains, does have long antichains (e.g. $\{(i,z): z\in\mathbb{Z}\}$ for any fixed $i\in I$). We can ask more subtle questions around both types of set; for example, is there an uncountable Boolean algebra in which every chain and antichain is countable? (See here.) In general, this sort of "uncountable Ramsey theory" takes us deep into combinatorial set theory and independence results.
Finally, note that if we drop choice things can get truly bizarre. It is consistent with $\mathsf{ZF}$ that there is an infinite Dedekind-finite set of real numbers with no least element (this occurs in Cohen's original model of $\mathsf{ZF+\neg AC}$); such a set, construed as a linear order in the usual way, is an uncountable linear order which satisfies your condition but doesn't have any infinite ascending or descending sequences! Generally some choice is needed to avoid serious silliness (on the other hand, sometimes we want to be silly ...).