# Weakly Compact Cardinal is Strong Limit

A question to Lemma 9.9 in Jech's Set Theory:

Every weakly compact cardinal is inaccessible.

I am working on the part that for $\kappa$ weakly inaccessible, $\kappa$ is strong limit.

Jech writes: That $\kappa$ is a strong limit cardinal follows from Lemma 9.4:
If $\kappa\leq 2^\lambda$ for some $\lambda<\kappa$, then because
$2^\lambda\not\rightarrow (\lambda^+)^2$, we have $\kappa\not\to(\lambda^+)^2$ and hence $\kappa\not\to(\kappa)^2$.

Why do we know then, that there is no $\lambda<\kappa$ with $\kappa\leq 2^\lambda$ and so $\kappa$ strong limit?

I don't see the connection between the partitions and this fact...

I hope someone feels like helping!

Best, Luca

• I don't understand the question. The quote you mention describes exactly the reason why there is no such $\lambda$. What part of the quote is giving you trouble? – Andrés E. Caicedo Aug 9 '13 at 23:20
• Perhaps the trouble is that you need to go from an implication [if $(\exists\lambda<\kappa)\,\kappa\leq2^\lambda$ then $\kappa\not\to(\kappa)^2$] to the contrapositive [if $\kappa\to(\kappa)^2$ then $\neg(\exists\lambda<\kappa)\,\kappa\leq2^\lambda$]. In other words, maybe the problem isn't with set theory but with logic. – Andreas Blass Aug 9 '13 at 23:42
• Sorry for being unclear! I think my problem is in understanding well the partition properties. In your comment Andreas, it's the 'then' in: if $\kappa\to(\kappa)^2$ then $\neg(\exists\lambda<\kappa)\kappa\leq 2^\lambda$ i think. Thank you already for the help! – Luca Aug 9 '13 at 23:55
• Thank you all.... it finally made click. Everything is clear now :D:D – Luca Aug 10 '13 at 0:16

## 1 Answer

1. If $\kappa=\lambda^+$ then $\kappa\leq2^{\lambda}$ due to Cantor's theorem. Therefore $\kappa$ is a limit cardinal.
2. Similarly, if $\kappa\leq2^\lambda$ for some $\lambda<\kappa$ we have that it's not weakly compact. Therefore $\kappa$ is a strong limit cardinal.
• Thank you already! Part 1 is clear, part 2: isn't a weakly compact cardinal also strong limit? I don't know how you followed: $\kappa$ is not weakly compact and therefore strong limit. sorry, and thanks! – Luca Aug 9 '13 at 23:57
• Luca, that depends on how you define a weakly compact cardinal. There are about gazillion definitions. In some we require that $\kappa$ is a strong limit, and from others we can deduce that immediately. – Asaf Karagila Aug 10 '13 at 0:11