Are the finite subsets of $\mathcal{P}(\mathbb{N})$, finite? I am trying to split up the finite subsets of $\mathcal{P}(\mathbb{N})$ into two disjoint groups $X \sqcup Y$ so that no two neighbouring sets are in the same group. (we introduce the term 'neighbouring': Two sets $A,B \subseteq \mathbb{N}$ are neighbouring, if $A$ is obtainable by adding an element to $B$, i.e. $A=B \cup \{c\}$ for a $c \not\in B$ or the other way around).
In my proof, I am trying to use the compactness theorem of propositional logic to find a partitioning for the finite subsets of  $\mathcal{P}(\mathbb{N})$, to deduce the fact that there exists a partitioning for $\mathcal{P}(\mathbb{N})$ as a whole.
Thus far, I could obtain a partition for the finite subsets $T:= \{1,...n\} \in \mathcal{P}(\mathbb{N})$ with $$X':= All \ subsets \ of \ T \ with \ an \ odd \ number \ of \ elements \\  Y':= All \ subsets \ of \ T \ with \ an \ even \ number \ of \ elements$$
so that $\mathcal{P}(T)=X' \sqcup Y'$.
Now, theoretically, from that partitioning, with the compactness theorem it should also follow that $\mathcal{P}(\mathbb{N})=X \sqcup Y$. But my two main concerns are:

*

*The compactness theorem is merely applicable in the sense of considering the finite subsets of the power set. But what if those subsets are, in itself, infinite? I.e., what if the subsets themselves do not have a finite number of elements.

*In light of 1, how would we be able to measure the "length" of those subsets? For instance, $\{even \ numbers\}$ and $\{1, \ even \ numbers\}$ would be neighbouring according to our definition. But the "length" of those sets would neither be odd or even, and would thereby not fit into either partition.

In this thread somebody has already tried to construct this partitioning, but used algebraic definitions. I feel like my proof thus far with the compactness theorem is right, but that last tiny bit is missing, or maybe I am overreacting and it is just fine as it is.
(In addition: My understanding of how the compactness theorem would apply here.
Our formula $\Phi$ is satisfied iff there exists a partitioning $\mathcal{P}(\mathbb{N})=X \sqcup Y$.
Now, applying the compactness theorem, we merely take the finite subsets of $\mathcal{P}(\mathbb{N})$, which we have definied as $\mathcal{P}(T)$, so that we have a partitioning for the finite subsets $\mathcal{P}(T)=X' \sqcup Y'$. Which is nothing else than a partition for the finite subsets (or, formulae) of our initial formula $\Phi$. And by definition of the compactness theorem, if every finite subset of the formula $\Phi$ is satisfied (we satisfy them by allocating them to either one of the disjunct groups), and we thus have a model for those formulae, we can conclude that it is also a model for $\Phi$ itself.)
Thank you so much in advance for any help!
Lin
 A: Note that "has a finite symmetric difference" is an equivalence relation on sets.  Partition the infinite subsets of $\Bbb N$ into equivalence classes.  Each class will be countably infinite, so there will be continuum many classes.  I claim you can assign one set from each class to $X$ or $Y$ as you wish, then all the rest of the sets in the class will be assigned by whether their symmetric difference from the first set has an even or odd (finite) number of elements.  For example, you might assign the even numbers to $X$.  You would be forced to assign the evens less $2$ to $Y$, the evens plus all the odds up to $11$ to $X$ and so on.  As the symmetric difference between any two sets of different equivalence classes you will not create a conflict with your neighbor relation however you assign the different classes.
The compactness argument goes like this:  You add a countable set of axioms "for any set of subsets of $\Bbb N$ with up to $c$ elements we can assign them to $X$ and $Y$ so that neighboring sets are in different groups" to your system, one for each value of $c$.  Your construction satisfies this for all finite subsets of the naturals.  You point out that any finite subset of these axioms has a finite maximum $c$ and your construction satisfies that subset, so it is a model of that set of axioms.  You then claim that by compactness the whole set of axioms has a model, which is a partitioning of all subsets of $\Bbb N$, finite or infinite.  I am not sure it works because the added axioms do not refer to the infinite subsets of $\Bbb N$
