# Is it consistent that $|[\kappa]^{<\kappa}| > \kappa$?

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 $$\kappa>\aleph_0$$?

• [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? ] – Qfwfq Feb 18 at 15:05
• @Qfwfq Yes, but more simply: "$\lnot P$ is not a theorem". – Alex Kruckman Feb 18 at 20:55
• @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). – Alex Kruckman Feb 18 at 21:01

## 1 Answer

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.

• Reading your answers is always like staring at a juggler. Everything looks so easy and natural, then when you try... the spell goes broken. – Ivan Di Liberti Feb 18 at 13:40
• 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... – Asaf Karagila Feb 18 at 13:42