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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.

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    $\begingroup$ You should assume that $\mu$ is a limit ordinal, since otherwise every $N$ is trivially internally approachable, by the sequence $\langle N\rangle$ of length $1$. Or, one can make longer sequences by padding with trivial objects first, and then $N$. $\endgroup$
    – Joel David Hamkins
    Commented Jun 15, 2010 at 20:58

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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 $N$.

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  • $\begingroup$ At first thank you very much for your answer. There is one thing though: You say that $N$ computes the length of $\mu$ by elementarity correctly. But why? Can't it be, for example that $(0,N_{0})$ is in $N$ whereas $\emptyset \notin N$? $N$ thinks that another $a /in N$ is the empty set but from the outside $a \ne \emptyset$? Thus only an $N$-ordinal is the length of our sequence $(N_{\alpha}, \alpha < \beta)$ and so the length is not correctly computed? $\endgroup$
    – Stefan Hoffelner
    Commented Jun 15, 2010 at 20:46
  • $\begingroup$ It should read 'the length of the sequence', not 'the length of $\mu$' in my comment. $\endgroup$
    – Stefan Hoffelner
    Commented Jun 15, 2010 at 20:49
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    $\begingroup$ If $N$ is elementary in $H(\theta)$, then it has the correct emptyset, the correct natural numbers and so on, since all these things are definable in $H(\theta)$. The length of the sequence $(N_\alpha| \alpha\lt\beta)$ is definable in $H(\theta)$, and so $N$ will compute this length correctly. $\endgroup$
    – Joel David Hamkins
    Commented Jun 15, 2010 at 20:52
  • $\begingroup$ Thank you again. I simply didn't realize that if $N$ thinks that an $a$, which is not empty in the outside, is the empty set, then, by elementarity $H(\theta)$ would think so too which is a contradiction. $\endgroup$
    – Stefan Hoffelner
    Commented Jun 15, 2010 at 21:02

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