Is there an infinite set $X$ such that for every isotone $f\colon[X]^\omega\to[X]^\omega$ there is a free decreasing sequence?

Question

(Pierre Gillibert asked me this question and I post it with his permission.)

Let $X$ be an infinite set, and $f\colon[X]^\omega\to[X]^\omega$. We say that $\{x_n\mid n<\omega\}\subseteq X$ is a free decreasing sequence (for $f$) if for all $n$, $x_n\notin f(\{x_k\mid k>n\})$.

Is there an infinite set $X$ such that for every isotone $f\colon[X]^\omega\to[X]^\omega$ there exists a free decreasing sequence? Is it at least consistent from assumptions such as $V=L$, large cardinals or strong forcing axioms?

Some observations:

1. It is clear that $X$ is uncountable, because otherwise $f(A)=X$ for all $A\in[X]^\omega$ would pose a counterexample.

2. If there is no such set of size $\kappa$, then there is no such example of size $\kappa^+$.

3. If $\kappa$ has uncountable cofinality, and there is no such set of size $<\kappa$, then there is no example of size $\kappa$, since we can "glue" counterexamples and use the fact that every countable set is bounded (this is in effect the same proof for the previous observation).

Answer 1

Here's a long, possibly unhelpful comment making use of a presumably excessive large cardinal assumption.

Suppose that $\lambda$ is an uncountable cardinal and that there is a nontrivial elementary embedding $j \colon L(V_{\lambda + 1}) \to L(V_{\lambda + 1})$ with critical point less than $\lambda$. So we are assuming that the large cardinal axiom I0 holds (see https://en.wikipedia.org/wiki/Rank-into-rank), with $\lambda$ here as the $\lambda$ there.

Note that (1) $j(\lambda) = \lambda$, (2), $j[\lambda]$ is in $L(V_{\lambda + 1})$ and (3) $[\lambda]^{\omega} \subseteq L(V_{\lambda + 1})$.

It seems that $L(V_{\lambda + 1})$ thinks that $\lambda$ is an $X$ as desired.

Claim : In $L(V_{\lambda + 1})$, for any $f : [\lambda]^{\omega} \to [\lambda]^{\omega}$ (not necessarily isotone) there exist an $\alpha < \lambda$ and an $X \subseteq \lambda$ of cardinality $\lambda$ such that $\alpha$ is not in $\bigcup f[[X]^{\omega}]$.

Applying the claim iteratively ought to let us build an independent family for any given $f$ in $L(V_{\lambda + 1})$. The output of our iterative construction is then in $L(V_{\lambda + 1})$, showing that $L(V_{\lambda + 1})$ thinks that $\lambda$ is as desired. Note that $L(V_{\lambda + 1})$ is not a model of Choice, so maybe this doesn't address the question.

Proof of claim. Let $f : [\lambda]^{\omega} \to [\lambda]^{\omega}$ in $L(V_{\lambda + 1})$ be given. Let $Z$ denote $j[\lambda]$. Let $\alpha$ be an element of $\lambda \setminus Z$ (for instance, the critical point of $j$). If $x$ is a countable subset of $Z$, then $x = j(y)$ for some $y \in [\lambda]^{\omega}$ (the pointwise $j$-preimage of $x$), so $j(f)(x) = j(f(y))$, which is contained in $Z$, so $j(f)(x)$ does not have $\alpha$ as a member. Then, in $L(V_{\lambda + 1})$, $Z$ is a subset of $\lambda$ of cardinality $\lambda$, and $\alpha$ is not in $\bigcup j(f)[[Z]^{\omega}]$. By the elementarity of $j$, then, we have the conclusion of the claim : there exist an $\alpha < \lambda$ and an $X \subseteq \lambda$ of cardinality $\lambda$ such that $\alpha$ is not in $\bigcup f[[X]^{\omega}]$.

Since I'm not using the isotone condition, then at least one of the following should hold : (1) I'm making a mistake (very likely), (2) the first sentence of the December 7 comment above uses more Choice than holds in $L(V_{\lambda + 1})$ or (3) I0 is inconsistent.