Onto functions from Power set of Naturals Let's define that a function (from $\mathcal{P}(\mathbb{N})$ to $\mathcal{P}(\mathbb{N})$, $\mathbb{N}$ meaning the set of natural numbers throughout this proof) is "OK" if:
$$\forall X,Y \in \mathcal{P}(\mathbb{N}). X \subset Y \leftrightarrow f(X) \subset f(Y)$$
while $X$ is a proper subset of $Y$, they are not equal (so as $f(X)$ and $f(Y)$).
Does there exist an "OK" and onto function $f: \mathcal{P}(\mathbb{N}) \setminus \{\mathbb{N}\} \rightarrow \mathcal{P}(\mathbb{N})$?
I tried disproving this claim, eventually getting stuck. I started with the assumption that $\mathbb{N}$ is in $Im(f)$, so there must be a subset of $\mathbb{N}$ (let's say $T$) so that $f(T) = \mathbb{N}$. $T$ is not $\mathbb{N}$ (because $\mathbb{N}$ is not in the domain), so there must be a natural number that is not inside $T$. and here I am stuck. Any suggestions?
Thanks!
 A: For clarity I write $X\subseteq Y$ for "$X$ is a subset of $Y$" and $X\subsetneqq Y$ for "$X$ is a proper subset of $Y$".
Theorem. There is no surjective function $f:\mathcal P(\mathbb N)\setminus\{\mathbb N\}\to\mathcal P(\mathbb N)$ such that $X\subsetneqq Y\iff f(X)\subsetneqq f(Y)$ for all $X,Y\in\mathcal P(\mathbb N)\setminus\{\mathbb N\}.$
The notation will be much simpler if I turn the question upside down, so that I can talk about finite sets instead of cofinite sets. Assume for a contradiction that there is such a function $f;$ then there is also a surjective function $g:\mathcal P(\mathbb N)\setminus\{\emptyset\}\to\mathcal P(\mathbb N)$ such that $X\subsetneqq Y\iff g(X)\subsetneqq g(Y)$ for all $X,Y\in\mathcal P(\mathbb N)\setminus\{\emptyset\}.$ Namely, you can easily verify that defining $g(X)=\mathbb N\setminus f(\mathbb N\setminus X)$ does the trick.
Claim 1. If $g(X)$ has $n$ proper subsets in $\mathcal P(\mathbb N),$ then $X$ has at least $n$ proper subsets in $\mathcal P(\mathbb N)\setminus\{\emptyset\}.$
Proof. Let $Y=g(X)$ and let $Y_1,\dots,Y_n$ be distinct proper subsets of $Y.$ Choose sets $X_1,\dots,X_n\in\mathcal P(\mathbb N)\setminus\{\emptyset\}$ with $g(X_i)=Y_i;$ then $X_1,\dots,X_n$ are distinct proper subsets of $X.$
Claim 2. Let $n$ be a positive integer. If $|X|=n,$ then $|g(X)|=n-1.$
Proof. If we had $|g(X)|\ge n,$ then $g(X)$ would have at least $2^n-1$ proper subsets; but this is impossible since $X$ has only $2^n-2$ nonempty proper subsets. Therefore $|g(X)|\le n-1.$ We prove by induction that $|g(X)|\ge n-1.$ For $n=1$ this is clear. If $n\gt1,$ then $X$ has an $(n-1)$-element subset $X'$; then $|g(X')|\ge n-2,$ and since $X'\subsetneqq X,$ we have that $g(X')\subsetneqq g(X)$ and so $|g(X)|\ge|g(X')|+1=n-1.$
Claim 3. For any set $X\in\mathcal P(\mathbb N)\setminus\{\emptyset\}$ we have $g(X)=\bigcup\{g(Y):Y\subseteq X,\ |Y|=2\}.$
Proof. We may assume that $|X|\ge3.$ On the one hand, if $Y$ is a $2$-element subset of $X,$ then $Y\subsetneqq X$ and so $g(Y)\subsetneq g(X).$ On the other hand, if $i\in g(X),$ then since $g$ is surjective we have $\{i\}=g(Y)$ for some $2$-element set $Y,$ and $Y\subsetneqq X$ since $g(Y)\subsetneqq g(X).$
Consider a $3$-element set $X=\{a,b,c\};$ let $g(X)=\{i,j\}.$ Without loss of generality we may assume that $g(\{a,b\})=g(\{a,c\})=i.$ Choose $d\notin X.$ Then we have $g(\{a,b\})=g(\{a,c\})\subsetneqq g(\{a,c,d\},$ but $\{a,b\}$ is not a subset of $\{a,c,d\}.$ We have arrived at a contradiction.
A: (In the following, $\subset$ means 'proper subset').
Since $f$ is to be surjective, exists some $T \in P(\mathbb{N})$ with $f(T) = \mathbb{N}$.
Suppose we have distinct naturals $x, y \notin T$; then $T \subset (T \cup \{y\}) \implies f(T) \subset f(T \cup \{y\}) \implies \mathbb{N} \subset f(T \cup \{y\})$, which is impossible since $\mathbb{N}$ cannot be a proper subset of any element of $P(\mathbb{N})$. Therefore $T$ must be of the form $\mathbb{N} - \{x\}$ for some natural $x$.
Next let $T'$ be any subset of $\mathbb{N}$ with $x \in T'$. If $f(T') \subset f(T)$ then $T' \subset T$; which is impossible since $x \notin T$. Therefore, $\forall T' \in P(\mathbb{N}) : x \in T' \implies f(T') = \mathbb{N}$.
But then let $T' = \{x, x+1\}$, and $T'' = \{x\}$. $T'' \subset T' \implies f(T'') \subset f(T') \implies \mathbb{N} \subset \mathbb{N}$ is a contradiction; so no such 'OK' $f$ exists.
