I'm reading "Classic Set Theory: For Guided Independent Study" and I'm trying to understand the proof of the recursion principle (I will just upload a picture since it's very long, you don't need to read it all I guess, but maybe it can help to answer the question, the part I don't understand is highlighted in yellow):
I don't understand how if $m∈n$, then $m∩n=m$. If for example $m=\{1,2,3\}$, and we need $m∈n$, this means that $n$ has to be at least of the form $n=\{\{1,2,3\},...\}$ and if $m∩n$ has to be equal to $m$ that means some of the elements of $n$ have to be $1,2,3$, but who tells us that they are? how is this implied from $m∈n$? Could you guys help me please? Thank you!
For starters, $m$ cannot be $\{1,2,3\}$, since that isn’t an element of $\Bbb N$. What you’re missing here is the actual definition of $\Bbb N$, whose elements are built up from $0$ by repeated applications of the successor operation $x^+=x\cup\{x\}$.
Your $\Bbb N$ contains $0$, $1=0^+=0\cup\{0\}=\{0\}$ (since by definition $0=\varnothing$),
$$2=1^+=1\cup\{1\}=\{0\}\cup\{\{0\}\}=\{0,\{0\}\}\;,$$
$$3=2^+=2\cup\{2\}=\{0,\{0\}\}\cup\{\{0,\{0\}\}\}=\big\{0,\{0\},\{0,\{0\}\}\big\}\;,$$
and so on. This definition is neatly designed so that for each $n\in\Bbb N$, $n=\{k\in\Bbb N:k<n\}$: each member of $\Bbb N$ is the set of smaller members of $\Bbb N$.
Now suppose that $m\in n$. This means that $m<n$, so
$$m=\{k\in\Bbb N:k<m\}\subseteq\{k\in\Bbb N:k<n\}=n\;:$$
$m$ is a subset of $n$ as well as an element of $n$. And of course it’s always true that if $A\subseteq B$, then $A\cap B=A$, so $m\cap n=m$.