Let $\kappa>0$ be a cardinal, and let $[\kappa]^{<\kappa}$ denote the collection of subsets of $\kappa$ having cardinality strictly less than $\kappa$. Is it consistent that $$[\kappa]^{<\kappa} > \kappa$$ for all cardinals $\kappa>\aleph_0$?

1$\begingroup$ [When you logicians say "it's consistent that $P$ holds" you mean that "Either $P$ is a theorem or $P$ is independent from the axioms", is my understanding correct? ] $\endgroup$ – Qfwfq Feb 18 at 15:05

2$\begingroup$ @Qfwfq Yes, but more simply: "$\lnot P$ is not a theorem". $\endgroup$ – Alex Kruckman Feb 18 at 20:55

3$\begingroup$ @Qfwfq Also, by the statement "$P$ is consistent" (with ZFC, say), we really mean "if $\text{ZFC}$ is consistent, then $\text{ZFC}+P$ is consistent". The reason for the hidden assumption is that if $\text{ZFC}$ is inconsistent, then no extension of it is consistent ($\lnot P$ is a theorem, because everything is). $\endgroup$ – Alex Kruckman Feb 18 at 21:01
Yes. First, let's just agree that $[\kappa]^{<\kappa}=\kappa^{<\kappa}$. One direction is immediate, in the other direction note that every function in $\kappa^{<\kappa}$ is an element of $[\kappa\times\kappa]^{<\kappa}$.
If $2^\kappa=\kappa^{++}$ for all successor cardinals, and there are no inaccessible cardinals, which is consistent by Easton's theorem then we can compute:
Either $\kappa$ is $\mu^+$, in which case $\kappa^{<\kappa}=\mu^\mu=\mu^{++}=\kappa^+$, or $\kappa$ is a limit cardinal in which case it is singular and $\kappa^{<\kappa}>\kappa$ anyway by König's lemma.

2$\begingroup$ Reading your answers is always like staring at a juggler. Everything looks so easy and natural, then when you try... the spell goes broken. $\endgroup$ – Ivan Di Liberti Feb 18 at 13:40

3$\begingroup$ Well. If it's any consolation, I don't feel like a juggler at all. If anything I feel as someone who doesn't know how to juggle who is trying to juggle six flaming swords, a baby and two chainsaws... $\endgroup$ – Asaf Karagila Feb 18 at 13:42