1
$\begingroup$

In Jech, Set Theory (1978), Solovay's Theorem is proved (Theorem 81, page 405). In the proof (page 407) we read:

Let $\kappa$ be a compact cardinal. If $\lambda>\kappa$ is an arbitrary cardinal, then we have $$\lambda^{<\kappa}\leq(\lambda^+)^{<\kappa}=\lambda^+$$ In particular, we have $\lambda^{\aleph_0}\leq \lambda^+$ for every $\lambda >\kappa$.

Up to here it is all clear. Now:

By Theorem 23 (or rather by Lemma 8.3), this implies that the singular cardinal hypothesis holds for every $\lambda>\kappa$.

Now can someone help me completing the proof? I recall that Lemma 8.3 says:

Let κ be a singular cardinal, let cf κ > ω, and assume that $λ^{cf\ κ} < κ$ for all $λ < κ$. If ${κ_α : α < cf\ κ}$ is a normal sequence of cardinals such that $\lim\ κ_α = κ$, and if the set $$\{α < cf\ κ : κ_\alpha^{cf κ_α} = κ_α^+\}$$ is stationary in $cf\ κ$, then $κ\ cf\ κ = κ^+$.

For $cf\ \lambda= \aleph_0$ the first quote gives us the result. For $cf\ \lambda>\omega$ we should apply the lemma. The problem is that I don't see how to build the normal sequence.

$\endgroup$
3
  • $\begingroup$ Why??? WHY??? WHY would you read that version of Set Theory, and not new one? The book had gone through two revisions! $\endgroup$
    – Asaf Karagila
    Commented Oct 23, 2017 at 15:13
  • $\begingroup$ Also, since I don't have that version of the book, I don't know if it's just a typo, or if the mistake is you misquoting from the book. But it is impossible that $\lambda<\kappa$ and that $(\lambda^+)^{<\kappa}=\lambda^+$. Specifically because $\lambda^+<\kappa$, so $\lambda^+<(\lambda^+)^{\lambda^+}<\lambda^{<\kappa}$. $\endgroup$
    – Asaf Karagila
    Commented Oct 23, 2017 at 15:16
  • $\begingroup$ Fixed, sorry. Numbering apart, the old is the same as the new in this part. In the new version, it is page 372-373. $\endgroup$
    – W. Rether
    Commented Oct 23, 2017 at 15:20

1 Answer 1

2
$\begingroup$

The set of all ordinals of countable cofinality is stationary in any ordinal of uncountable cofinality.

In combination of Lemma 8.3 this implies that SCH for countable cofinality implies SCH anywhere.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .