So, I don't know if this certain type of set has a name in English. If it has, I'll be glad if someone will edit my post.
Definition: Let $A$ be a set and $f: P(A)\setminus \emptyset \rightarrow A$ a choice function. A well-ordered set $(B,R)$ is called proper if $B \subseteq A$ and for every $b \in B: f(A\setminus B_b)=b $. ($B_b$ is the initial segment with $b$ being upper bound)
Question: Let there be $\mathbb N$ and $S=P(\mathbb N)\setminus \emptyset$.
We'll define a choice function $f: S \rightarrow \mathbb N$ as following: For every finite set $X$, $f(X)=max(X)$, that is - maximal value with regular $<$ relation on $\mathbb N$. For every infinite set $X$, $f(X)$ is the 10th element of the set (10th according with the regular $<$ relation on $\mathbb N$).
Find the proper subsets and what's the order on $\mathbb N$ you get as a result.
I tried solving this for a very long time. Any full solution would be very appriciated.
The ordering that you get is
$$9<10<11<\cdots 8<7<6<5<4<3<2<1<0$$ the proper sets are just the initial segments.
This is obtained as follows $$f(\mathbb{N})=9$$ $$f(\mathbb{N} - \{9\})=10$$ $$f(\mathbb{N} - \{9,10\})=11$$ etc $$f(\{0,1,2,3,4,5,6,7,8\})=8$$ $$f(\{0,1,2,3,4,5,6,7\})=7$$ and so on.