Countable elementary submodels I'm having some trouble understanding elementary submodels.
Let $H_\chi$ be the set of all sets which are hereditarily of cardinality $<\chi$. Let $\textbf{N}=(N,\in)$ be a countable elementary submodel of $(H_{\chi} ,\in)$ Moreover let $p\in N$ and $q\in H_{\chi}$ such that $\lvert p\triangle q\rvert$ is finite. Under what sort of circumstances is $q\in N$?
 A: The following are equivalent:


*

*$(i)$ $q\in N$,

*$(ii)$ $p\Delta q\subseteq N$,

*$(iii)$ $p\Delta q\in N$.
$(ii)\rightarrow(iii)$ is easy, since by assumption $p\Delta q$ is finite - it's easy to show that the union of finitely many elements of $N$ is in $N$.
For $(iii)\rightarrow(i)$, note that $N$ contains an element $r$ which $N$ thinks satisfies $r=p\Delta(p\Delta q)$ - by elementarity, we in fact have $r=p\Delta(p\Delta q)$, but this is just $q$.
For $(i)\rightarrow (iii)$, if $q\in N$ then $N$ contains some element $s$ which $N$ thinks is $p\Delta q$ - so by elementarity, $s=p\Delta q$.
Finally, for $(iii)\rightarrow (ii)$, this follows from the following more general fact:

Suppose $N$ is a countable elementary submodel of $H_\chi$ and $X\in N$ is countable. Then $X\subseteq N$.

Pf: By elementarity, $N$ contains some $f$ which $N$ thinks is a surjection from $\omega^N$ to $X$. But $\omega^N=\omega$, so by elementarity $f$ is in fact a surjection from $\omega$ to $X$. But then each $x\in X$ is of the form $f(n)$ for some $n\in\omega$, and $\omega\subseteq M$ - so for each $n\in\omega$, $f(n)\in N$.

Note that these conditions aren't vacuous. For example, fix $p\in N$ and $x\in H_\chi\setminus N$. Then $q:=p\cup\{x\}\not\in N$.
