I just read for the first time the definition of an internally approachable set, which says:


A set $N$ is internally approachable (I.A.) of length $\mu$ iff there is a sequence $(N\_{\alpha} : \alpha < \mu)$ for which the following holds: $N=\bigcup\_{\alpha< \mu} N\_{\alpha}$ and for all $\beta < \mu$ $( N\_{\alpha} : \alpha < \beta ) \in N$.


Now if $N \prec (H(\theta), \in, < )$ is I.A. of length $\mu$. Is it true that


(a) If $\alpha < \mu$ then $\alpha \in N$ 


(b) If $\alpha < \mu$ then $N\_{\alpha} \in N$ ?


This is trivial if $N$ is transitive, and I'm quite sure that both (a) and (b) hold but I need a good argument.



Some consequences of internally approachable structures
Yes, both (a) and (b) follow from your definition without assuming that $N$ is transitive. You say that for every $\beta\lt\mu$ the sequence $\langle N\_\alpha | \alpha\lt \beta\rangle$ is in $N$. This implies that $\beta$ is in $N$, since $\beta$ is the length of this sequence and $N$ computes this length correctly by elementarity. Thus, every $\beta\lt\mu$ is in $N$ and so (a) holds (renaming $\alpha$ to $\beta$). It now follows that (b) also holds, assuming that $\mu$ is a limit ordinal, since once we have the sequence of length $\alpha+1$ in $N$, we may evaluate this sequence at $\alpha$ to deduce that $N\_\alpha$ is in $N$.


