I am trying to understand/reconstruct the proof given by my Professor addressing the existence of natural numbers. However there is one step in particular I don't understand and the more I think about it, the less I do.
Definition: A System of natural Numbers is the triple (N,0,S) where $N$ denotes a Set, $0 \in N$ an element and $S: N \to N$ a mapping such that
1) $S$ is injective
2) $0 \notin S(N)=Im(S)$
3) If $M \subset N$ with $0 \in M$ and $S(M) \subset M$ then $M=N$
Proposition: Let $U$ be an infinite Set. Then there exists a System of natural numbers $(N,0,S)$ with $N \subset U$
Proof: Let $U$ be an arbitrary infinite set, by definition of infinite sets (Dedekind) there exists a mapping $\varphi: U \to U$ which is injective but not bijective. In the following let $\varphi$ denote such a mapping.
Choose $u \in U \setminus \varphi(U)$ and consider the set $m:= \lbrace N' \in P(U) \mid u \in N', \varphi(N') \subset N'\rbrace$. It is clear that $m \neq \emptyset$ because $U \in m$.
Lets define:
\begin{align}N:=& \bigcap_{N' \in m}N' \tag{i} \\ 0:=&u \tag{ii} \\ S:&N \to N, \ S(n):= \varphi(n) \tag{iii}\end{align}
Here comes the crucial step I don't understand, my Professor said the definition of $S$ is informal, more formally one would have to define $$S:= \varphi_{ \mid N}:N \to U $$ and then show that $\varphi(N) \subset N$ such that $S$ is well defined. This makes sense to me, but my Professor now does the following: $$\varphi(N) \subset N, \text{ because } N= \bigcap_{N' \in m} N' $$ and this has me lost for hours now. It is clear that $N \subset U$ because $U \in m$ and the above definition intersects over all $N' \in m$, but I don't see that these 'intersections' should inherit the attributes defined in the set $m$
The intersection doesn't automatically inherit the properties of the sets being intersected in general; you have to prove it does in this case. Specifically:
Let $x \in N$. Then for $N' \in m$, we have $x \in N'$, so by definition of $m$, $\varphi(x) \in N'$. Since this holds for all $N' \in m$, we conclude that $\varphi(x) \in N$.