Split all subsets of $\mathcal{P}(\mathbb{N})$ into 2 groups $A \dot\cup B =\mathcal{P}(\mathbb{N})$ s.t. no 2 neighboring sets are in the same group So I'm trying to come up with a proof of the above action being possible or not, using the compactness theorem from logics (a set of first-order sentences has a model if and only if every finite subset of it has a model) on this one but I'm not quite sure how to split up the finite subsets of $\mathbb{N}$ to make the condition true.
(Note: Neighboring here means that for instance $X,Y \subseteq \mathbb{N}$ are neighboring if you can get $X$ by adding an element to Y, i.e. $X=Y \cup \{c\}$ for a c $\not\in Y$ or the other way around. $\{2,41\}$ and $\{0,2,41\}$ would be neighbors but not $\{3,4\}$ and  $\{3,5\}$
Note 2: $\dot\cup$ is the disjoint union
Note 3: Referring to all possible subsets of $\mathcal{P}(\mathbb{N})$. @Gae. S. provided a very nice answer using the axiom of choice, but I am looking for a way to prove it with the compactness theorem of logics which is in a certain way similar to the axiom of choice.
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I'd really appreciate some help on this :) Thanks so much!
 A: This is essentially a variation of the dwarfs-and-hats problem, or at least the same use of the axiom of choice.
Consider the equivalence relation on $\mathcal P(\Bbb N)$ defined by $$X\equiv Y\iff \lvert X\triangle Y\rvert<\aleph_0$$
where $X\triangle Y=(X\setminus Y)\cup(Y\setminus X)$ and $\lvert X\rvert$ is the cardinality of $X$. Consider the quotient set $M=\mathcal P(\Bbb N)/_\equiv$ and a map $f:M\to \mathcal P(\Bbb N)$ assigning to each equivalence class a representative. Id est, a map such that $f([X]_\equiv)\triangle X$ is a finite set for all $X\subseteq \Bbb N$.
Then, define the two sets by $$A=\{X\in\mathcal P(\Bbb N)\,:\, \lvert X\triangle f([X]_\equiv)\rvert\text{ is odd}\}\\ B=\{X\in\mathcal P(\Bbb N)\,:\, \lvert X\triangle f([X]_\equiv)\rvert\text{ is even}\}$$
A: Let $p_S$ be a propositional variable for each $S \subset \mathbb N$.
Take for axioms the uncountably many sentences:
$$ p_S \not\equiv p_T $$
whenever $S,T \subset \mathbb N$ are neighboring, i.e. their symmetric difference is a singleton set.
We show that any finite subset $\mathbf S$ of those sentences has a model (and so is consistent).
Consider the graph $\mathcal G$ whose vertices are the finitely many sets corresponding to any of the sentential variables $p_S$ appearing in $\mathbf S$, and whose edges are between pairs of neighboring sets.
Lemma The graph $\mathcal G$ is bipartite.
Proof It suffices to show that $\mathcal G$ has no cycles of odd length.
Let $S_0 \sim S_1 \sim \ldots \sim S_n$ be a path in $\mathcal G$.
Since each edge in a cycle signifies the addition or removal of an element from one set to obtain the next set, any two nodes in a path will have a finite symmetric difference.  By induction these symmetric differences satisfy:
$$ | S_i \Delta S_j | \equiv |i-j| \bmod 2 $$
Applying this to a cycle of length $n$, where necessarily $S_0 = S_n$, we see that the cycle length is even. QED
Finally we interpret $\mathcal G$ as the model for the sentences $\mathbf S$.  Let all the sentential variables $p_S$ for sets $S$ in one part of $\mathcal G$ be true and those variables for sets in the other part be false.  Since the axioms merely require that two propositional variables which correspond to neighboring subsets of $\mathbb N$ have opposing truth values, all the axioms in $\mathbf S$ are satisfied in this model.
It follows by the compactness theorem for the propositional calculus that there exists a model, and therefore a consistent interpretation, of the full set of (uncountably many) axioms.  This interpretation bipartitions $\mathcal P(\mathbb N)$ into those subsets $S$ of $\mathbb N$ corresponding to true values of $p_S$ and those corresponding to false values, just as desired.
A: The way I understand it, two sets are neighbors if one has $n$ elements and the other has $n + 1$ elements.
You could put all the sets with an odd number of elements into $A$, and all the sets with an even number of elements into $B$.
