Concerning increasing functions defined on (limit) ordinals.

Question

This is related to this question. Let $(X,\leq)$ be an ordered set (Edit: I am not assuming that the order $\leq$ is linear; see my fake and friendly "flame war" with Brian M. Scott in his answer below), and let $\kappa$ be a limit ordinal (endowed with the usual order, or course). Consider the condition "every increasing function $\require{enclose}\enclose{horizontalstrike}[mathcolor="red"]{\color{black}{f:X\to\kappa\ }}\ f:\kappa\to X$ is eventually constant", which I call "ascending $\kappa$-chain condition on $X$" (probably a standard naming).

For $\kappa=\omega$, this is equivalent (under the axiom of dependent choice) to the condition "every nonempty subset of $X$ has a maximal element". At first I thought that the ascending $\kappa$-chain conditions were stronger for $\kappa>\omega$ , but later I realized that this is not the case, precisely by the fact that $\kappa$ is larger than $\omega$: in fact you can impose that every $\kappa$-ascending chain stabilizes, yet this perhaps occurs after the $\omega$-th step, so you can have examples of strictly increasing $\omega$-chains.

Since the ascending $\omega$-chain condition is equivalent to such "strong" statement (involving maximality), I was quite sure that the converse was indeed true. If ascending $\omega$-chain condition holds, how to prove that any $\kappa$-ascending chain $(a_\lambda)_{\lambda\in\kappa}$ on $X$ stabilizes?

If $\kappa$ has a denumerable cofinal subset, say $S$, then we are done, because the subchain $(a_\lambda)_{\lambda\in S}$ stabilizes, which forces the original chain to stabilize. But what about the case in that $\kappa$ has no denumerable cofinal subset? what can be said in this case? I take this opportunity to repeat the question I posted before: what is an (interesting) equivalent to the ascending $\kappa$-chain condition? (for this question please post your answer on the corresponding post). Thanks in advance.

EDIT

The characterization of the $\omega$-ascending condition works on any ordered set, that is, for any ordered set $\boldsymbol{X}$, the ascending $\omega$-chain condition holds on $X$ iff every nonempty subset of $X$ has a maximal element. I want to know about equivalent characterizations of the $\kappa$-ascending chain condition for arbitrary ordered sets (that is, any $\kappa$-ascending chain on any ordered set stabilizes iff ???). Trevor Wilson's answer settles the question only on the case that $X$ is an ordinal as well.

Minor stylistic commentary: "order-preserving" is far more canonical than "increasing", but I think that the former is less appealing in this case.

Major stylistic commentary: I always mean $X$ to be a partially ordered set (in view of Andrés Caicedo and Brian M. Scott comments).

BIG EDIT

I am terribly sorry to all: at the first paragraph I meant "$f:\kappa\to X$" instead of $f:X\to\kappa$". I swear, I always had in mind $f:\kappa\to X$. Thanks to Brian M. Scott to indirectly pointing out my mistake through his digression about the ambiguity of the (flawed) definition in the case that $X$ is not a linearly ordered set.

Answer 1

If every weakly increasing function $X \to \omega$ is eventually constant, it does not follow that every weakly increasing function $X \to \omega_1$ is eventually constant. For example, let $X$ be $\omega_1$ with the usual ordering. Every weakly increasing function $\omega_1 \to \omega$ is eventually constant, but the identity function $\omega_1 \to \omega_1$ is not.

Regarding your second question, if $X$ happens to be an ordinal, say $X = \xi$, then an equivalent condition to the statement "every weakly increasing function $X \to \kappa$ is eventually constant" is the statement "$\text{cof}(\xi) > \kappa$", where $\text{cof}(\xi)$ denotes the cofinality of $\xi$.

Answer 2

$\newcommand{\cf}{\operatorname{cf}}$In your terms a linear order $\langle X,\le\rangle$ has the $\kappa$-ACC iff $\cf X>\kappa$, where $\cf X$ is the cofinality of $X$, i.e., the least cardinal $\lambda$ for which there is a strictly increasing $\lambda$-sequence $\langle x_\xi:\xi<\lambda\rangle$ in $X$ such that for each $x\in X$ there is a $\xi<\lambda$ such that $x\le x_\xi$.

Let $\kappa$ be an uncountable regular cardinal, and let $\langle X,\le\rangle$ be a linear order such that every weakly order-preserving function $f:X\to\kappa$ is eventually constant. Suppose that $\cf X=\lambda$, and let $\langle x_\xi:\xi<\lambda\rangle$ be a strictly increasing cofinal $\lambda$-sequence in $X$. If $\lambda\le\kappa$, define

$$f:X\to\kappa:x\mapsto\min\{\xi<\lambda:x\le x_\xi\}\;;$$

clearly $f$ is weakly order-preserving and not eventually constant. Thus, $\lambda>\kappa$.

Now suppose that $\langle X,\le\rangle$ is a linear order of cofinality $\lambda>\kappa$, and let $\langle x_\xi:\xi<\lambda\rangle$ be as before. Let $f:X\to\kappa$ be weakly order-preserving. Then $\langle f(x_\xi):\xi<\lambda\rangle$ is a non-decreasing $\lambda$-sequence in $\kappa$ and so must be eventually constant. Fix $\eta<\lambda$ and $\alpha<\kappa$ such that $f(x_\xi)=\alpha$ for $\eta\le\xi<\lambda$. Then $f(x)=\alpha$ for all $x\in X$ such that $x_\eta\le x$, so $f$ is eventually constant.

In other words, Trevor Wilson’s answer generalizes readily to arbitrary linear orders.

Added: To deal with arbitrary partial orders, you need to clarify what you mean by eventually constant: on one natural interpretation it’s not true that a partial order has the $\omega$-ACC iff every non-empty subset has a maximal element.

Let $X=\{\langle m,n\rangle\in\omega\times\omega:n\le m\}$ equipped with the partial order $\langle m,n\rangle\preceq\langle r,s\rangle$ iff $m=r$ and $n\le s$. For $m\in\omega$ let $C_m=X\cap\big(\{m\}\times\omega\big)$. If $\varnothing\ne S\subseteq X$, then $S$ has a maximal element for each $m\in\omega$ such that $S\cap C_m\ne\varnothing$, but the function $$f:X\to\omega:\langle m,n\rangle\mapsto n$$ is weakly order-preserving and has unbounded range.

Do you want to define $f$ is eventually constant on $X$ to mean that for each $x\in X$ there is a $y\in x$ such that $x\le y$ and $f$ is constant on $\{z\in X:y\le z\}$?