Strong limit cardinals below the first inaccessible This is from Thomas Jech's "Set Theory" - Third Millennium Edition - page 95 (Section on Mahlo Cardinals). If k is the least inaccessible, all strong limits  below it (if there are any) have to be singular. Understood. But he says the set of all these must be closed and unbounded. Why? In the extreme case, why can't k be the least strong limit cardinal which also happens to be regular? So the set of strong limits below k is empty. Alternatively, there are such strong limits, but these don't reach all the way up to k?
 A: Short version: The situations you describe in your last sentence would indeed contradict the claim, but it can't occur; meanwhile the situation described in your third-to-last sentence has no bearing on the problem since regularity isn't part of the definition of strong limit-ness. The claim is true, and is proved by reasoning about the combinatorics of strong limit cardinals.

Long version: First of all, if $\kappa$ is the least inaccessible then by definition it is the least regular strong limit cardinal; this is what "inaccessible" means. (OK fine it's what "strongly inaccessible" means, but that's what's usually meant when we say "inaccessible.") But this has no bearing on the claim: the claim is that 
$$S=\{\lambda<\kappa:\lambda\mbox{ is a strong limit cardinal}\}$$
is club below $\kappa$, and you'll note that the word "regular" does not appear in the definition of $S$. Strong limits are not in general regular. For example, the first strong limit cardinal is $$\beth_\omega=\sup\{2^{\aleph_0}, 2^{2^{\aleph_0}}, 2^{2^{2^{\aleph_0}}}, ...\}$$ (do you see why?) and this has cofinality $\omega$. 
The key points here are the following:

Claim 1: A limit of strong limit cardinals is also a strong limit cardinal.

Proof sketch: if $\lambda$ is a limit of strong limit cardinals, clearly $\lambda$ is a limit cardinal. Now suppose $\mu<\lambda$; we want to show that $2^\mu<\lambda$. Since $\lambda$ is a limit of strong limits, there is some strong limit $\beta$ with $\mu<\beta<\lambda$; do you see how to proceed from here? $\quad\Box$
This shows that $S$ is closed. Now, why is $S$ unbounded below $\kappa$? Well, suppose $\sup(S)=\lambda<\kappa$. By Claim 1, $\lambda$ must be a strong limit cardinal (why?). Now consider:

Claim 2: If $\lambda<\kappa$, then $\beth_\omega(\lambda)<\kappa$.

This follows from the strong limit-ness and regularity (or rather, uncountable cofinality) of $\kappa$.

Claim 3: For all $\lambda$, the cardinal $\beth_\omega(\lambda)$ is a strong limit cardinal.

This is proved in the same way that we prove that $\beth_\omega$ is a strong limit cardinal.
Putting claims $2$ and $3$ together we get a contradiction with the assumption $\sup(S)<\kappa$; this gives that $S$ is unbounded below $\kappa$, and so we're done.
