small chains guarantee existence of big independent set? I can't figure out whether the hypothesis below holds. If you know the answer (or else the decidability), I would love to get a hint towards understanding it.
Definition. If $(P,\prec)$ is a partially ordered set and $I\subset P$, then $I$ is called an independent subset of $P$ if for all $x,y\in I$ we have $x\prec y$ nor $y\prec x$. For the sake of clarity, this means: not ($x\prec y$ or $y\prec x$).
Hypothesis. Assume GCH, let $\kappa\geq\mu^{++}$ be regular, and let $\prec$ be a partial order on $\kappa$ such that every $\prec$-chain has cardinality at most $\mu$. Then $\kappa$ has an independent subset of cardinality $\kappa$.
I included GCH because of the following counterexample for $\kappa=2^\mu$:
Let $\prec'$ be the lexicographic ordering on $2^\mu$, let $<$ be a wellordering of $2^\mu$ and set ${\prec}={\prec'}\cap{<}$. Every $\prec$-chain is a $\prec'$-chain so it has cardinality at most $\mu$, but every ($\prec$-)independent subset is also a $\prec'$-chain and thus has cardinality at most $\mu$.
Proof for $\mu<\omega$. By induction. The cases $\mu=0$ and $\mu=1$ are trivial.
$\mu\geq2$: Let $B=\{\zeta\in\kappa:\exists\lambda\in\kappa:\zeta\prec\lambda\}$. Then $\kappa\setminus B$ is clearly an independent subset. We may therefore assume that $|B|=\kappa$. We apply the induction hypothesis with $B,\prec|B$ replacing $\kappa,\prec$ respectively to find a $\prec$-chain $C=\{c_1\prec\dots\prec c_{\mu-1}\}$ in $B$ of length $\mu-1$. By definition of $B$ there is $\zeta\in\kappa$ with $c_{\mu-1}\prec\zeta$; $C\cup\{\zeta\}$ is a $\prec$-chain in $\kappa$ of cardinality $\mu$.
 A: I found a weaker statement which could be relatively superficial, but which is still quite satisfying for me at the moment. If you want, feel free to check the validity of the proof.
Lemma. Let $\kappa>2^\mu$ and let $\prec$ be an asymmetric relation on $\kappa$ such that every subset that is totally ordered by $\prec$ has cardinality $<\mu$. Then there is an independent subset $I\subset\kappa$ of cardinality $\mu$ (independent again means that for all $x,y\in I$, $x\prec y$ nor $y\prec x$).
Proof. The idea is that we keep splitting up the set $\kappa$ so that elements that have not been separated, agree (regarding $\prec$) on a certain growing set. We do this $\mu$ times, and then there have to be some unseparated elements left (this will be called $A_\delta$ at the end of the proof); the growing set below this, which will be the image of $\phi$, is then homogeneous enough to forcedly contain either a chain or an independent set of cardinality $\mu$.
Let $c$ be a choice function on $\mathcal{P}\kappa\setminus\{\emptyset\}$. For every $\zeta\leq\mu$ and $\alpha\in3^\zeta$ we define a subset $A_\alpha\subset\kappa$ such that


*

*$A_\emptyset=\kappa$;

*$A_\alpha\subset A_\beta$ for $\alpha\supset\beta$;

*$A_\alpha\cap A_\beta=\emptyset$ for $\alpha,\beta\in3^\zeta(\alpha\neq\beta)$;

*$|A_\beta\setminus\bigcup_{\beta\subset\alpha\in3^\zeta}A_\alpha|\leq2^\mu$ for all $\beta\in3^\xi$ and $\xi\leq\zeta$;

*$(x\prec c(A_\beta)\Leftrightarrow \alpha(\xi)=0)$ and $(c(A_\beta)\prec x\Leftrightarrow \alpha(\xi)=2)$ for all $\xi<\zeta$, $\beta\in3^\xi$, $\alpha\in3^\zeta$, $\beta\subset\alpha$, $x\in A_\alpha$.


Suppose that we have defined $A_\beta$ where $\beta\in3^\zeta$. We write $\beta[r]=\beta\cup\{(\zeta,r)\}$ for $r\in3$. If $A_\beta=\emptyset$ we set $A_{\beta[r]}=A'_\beta=\emptyset$ for all $r$. Otherwise, let $A'_\beta=A_\beta\setminus\{c(A_\beta)\}$, $A_{\beta[0]}=\{a\in A'_\beta:a\prec c(A_\beta)\}$, $A_{\beta[2]}=\{a\in A'_\beta:c(A_\beta)\prec a\}$ and $A_{\beta[1]}=A'_\beta\setminus(A_{\beta[0]}\cup A_{\beta[2]})$.
If $\zeta$ is a limit ordinal and $\alpha\in3^\zeta$, let
$$
A_\alpha=\bigcap_{\xi<\zeta}A_{\alpha|\xi}.
$$
This completes the definition of the sets $A_\alpha$.
It is clear that 2 and 3 hold.
Let's verify 4. Let $\xi$ and $\beta$ be fixed. It is trivial for $\xi=\zeta$. Suppose that it holds for certain $\zeta$. Then
$$
\bigcup_{\beta\subset\alpha\in3^{\zeta+1}}A_\alpha=\bigcup_{\beta\subset\alpha\in3^\zeta}(A_{\alpha[0]}\cup A_{\alpha[1]}\cup A_{\alpha[2]})=
\bigcup_{\beta\subset\alpha\in3^\zeta}A'_\alpha
$$
so
$$
\left|A_\beta\setminus\bigcup_{\beta\subset\alpha\in3^{\zeta+1}}A_\alpha\right|\leq\left|A_\beta\setminus\bigcup_{\beta\subset\alpha\in3^\zeta}A_\alpha\right|+|3^\zeta|\leq2^\mu.
$$
Now if $\zeta$ is a limit ordinal, we claim that
$$
\bigcup_{\beta\subset\alpha\in3^\zeta}A_\alpha
$$
contains $A_\beta\setminus\{c(A_\gamma):\eta<\zeta,\gamma\in3^\eta(A_\gamma\neq\emptyset)\}$.
If $x$ is an element of this last set, we apply the induction hypothesis on this last claim to find for each $\xi\leq\eta<\zeta$ an $\alpha_\eta$ with $\beta\subset\alpha_\eta\in3^\eta$ and $x\in A_\eta$. 2 and 3 yield $\eta<\eta'\implies\alpha_\eta\subset\alpha_{\eta'}$. For $\delta:=\bigcup_\eta\alpha_\eta$ we have
$$
x\in\bigcap_{f\in\zeta^{(3^\zeta)}}A_{\delta|f(\delta)}\subset\bigcap_{f\in\zeta^{(3^\zeta)}}\bigcup_{\beta\subset\alpha\in3^\zeta}A_{\alpha|f(\alpha)}
=\bigcup_{\beta\subset\alpha\in3^\zeta}\bigcap_{\eta<\zeta}A_{\alpha|\eta}=\bigcup_{\beta\subset\alpha\in3^\zeta}A_\alpha.
$$
This proofs 4.
To verify 5, note by 2 that $x\prec c(A_\beta)\Leftrightarrow x\in A_{\beta[0]}$. By 2 and 3, this is equivalent to $\alpha(\xi)=0$. The second statement is similar.
We now construct the subset $I$. By 1 and 4,
$$
\left|\bigcup_{\alpha\in3^\mu}A_\alpha\right|=\kappa.
$$
Thus there exists $\delta\in3^\mu$ such that $A_\delta\neq\emptyset$. Let $\phi(\zeta)=c(A_{\delta|\zeta})$ for $\zeta\leq\mu$. There are three cases.


*

*$|\delta^{-1}\{0\}|=\mu$: let $C=\phi(\delta^{-1}\{0\})$. We prove that $C$ is totally ordered by $\prec$ and of cardinality $\mu$ by showing that $\phi:(\delta^{-1}\{0\},<)\rightarrow(\kappa,\prec)$ is order-reversing. Let $\xi,\zeta\in\delta^{-1}\{0\}$ with $\xi<\zeta$ and note that $\phi(\zeta)\prec\phi(\xi)$ by 5.

*$|\delta^{-1}\{2\}|=\mu$: this case is similar to case 1.

*$|\delta^{-1}\{1\}|=\mu$: let $I=\phi(\delta^{-1}\{1\})$. By the same arguments as in the previous cases, $I$ is an independent subset. To show that $|I|=\mu$, let $\xi,\zeta\in\delta^{-1}\{1\}$ with $\xi<\zeta$ and note by 2 that $\phi(\zeta)\in A_{\delta|\xi+1}\subset A'_{\delta|\xi}$ so $\phi(\zeta)\neq\phi(\xi)$.


The proof is complete.
