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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:

  1. 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.
  2. 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

Lin Shao
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    Are you trying to partition the finite subsets of $P(\Bbb N)$ or the finite elements of $P(\Bbb N)$? It sounds like you are partitioning the finite subsets of $\Bbb N$. Then $X$ can be all the finite subsets of $\Bbb N$ with an odd number of elements and $Y$ can be the finite subsets of $\Bbb N$ with an even number of elements. The infinite subsets don't go in either one so you don't have $P(\Bbb N)=X \cup Y$ – Ross Millikan May 09 '21 at 14:42
  • Finite subsets of $P(\Bbb N)$ are things like ${{1},{2,3,17},{1,2,3}}$. They are not subsets of $\Bbb N$ at all. That was the point of my question. I think your proposed partition works either way for the finite subsets because you just count the number of elements (whatever they are). You still will not have partitioned $P(\Bbb N)$ because you have not assigned the infinite subsets anywhere. – Ross Millikan May 09 '21 at 14:57
  • Oh right I'm sorry, my brain had a short circuit right there... well, obviously, the odd and the even numbers would be subsets of $\mathbb{N}$ like you said. The aim here is to try and find a partitioning for the finite subsets of $\mathbb{N}$ so that $\mathcal{P}(T) = X' \sqcup Y'$ (which we have) to then prove that there exists a partitioning (also in odd and even) for $\mathcal{P}(\mathbb{N}) = X \sqcup Y$. – Lin Shao May 09 '21 at 15:04
  • I think for the infinite subsets you may need to proceed by transfinite induction. Pick one infinite subset, say the evens and put it in $X$, say. Then put subsets in $X$ and $Y$ as they are forced, so the evens less one member would go in $Y$ as would the evens plus one odd. Continue until you run out of subsets that you can place. You won't have placed the odds yet. You would like to show that you can place them either in $X$ or $Y$ without causing a conflict. I believe that is true. – Ross Millikan May 09 '21 at 15:10
  • I had that at the back of my mind since I've heard of transfinite induction in the past, however, I have no clue how it works unfortunately (and we haven't been taught transfinite induction so far, anyway) - that's why the aim is to prove it by the compactness theorem – Lin Shao May 09 '21 at 15:12
  • You can use the fact that "has a finite symmetric difference" is an equivalence relation on sets. Partition the infinite subsets of $\Bbb N$ into equivalence classes under this relation. I believe you can then place one set of each class into $X$ or $Y$ at will, then you will know where to place all the others and you will not have a conflict. – Ross Millikan May 09 '21 at 15:13
  • Thank you! I believe that your idea is just the answer that I've mentioned in the other thread. I've come to understand that one now thanks to you, nevertheless, I don't see how one could incorporate that into a compactness theorem solution – Lin Shao May 09 '21 at 15:20
  • Is the point of this problem, perhaps, that "even--odd" solution (in the answer below, as well as in the answer to the linked post; are these posts coming from some class?) depends on the axiom of choice, whereas the compactness theorem is strictly weaker than the axiom of choice? – Lee Mosher May 09 '21 at 15:50
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    Most of what you have written just doesn't make sense at all; you are completely on the wrong track about how to use the compactness theorem here (though you are on track to a different solution that doesn't use compactness at all). I would suggest backing up and thinking very carefully about precisely what the compactness theorem says and how it could even potentially apply here. The compactness theorem gives a criterion for a set of propositions to be satisfiable, so what precisely would those propositions be here? What would be the propositional variables from which they are built? – Eric Wofsey May 09 '21 at 16:01
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    @LinShao: You still have not updated the question in light of the comments. It is important that you want to partition all the subsets of $\Bbb N$, finite and infinite. It is important that we are talking of subsets of $\Bbb N$, not subsets of $P(\Bbb N)$. Getting the question right is the first step to a solution. -1 – Ross Millikan May 10 '21 at 02:49
  • I have posted an Answer to the linked Question making use of the compactness theorem for the propositional calculus. Please take a look. Thanks for bringing it to my attention. – hardmath May 10 '21 at 15:42

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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 $\Bbb N$

Ross Millikan
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  • I think you are reproducing the "algebraic" argument given in the Question's linked thread, so it likely is not going to be easy for the OP here to follow. I suspect there is an interpretation of the problem amenable to applying the compactness theorem, as that seems to be an exercise from some textbook used by both this student and the one who posted the linked Question. – hardmath May 10 '21 at 02:16
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    @hardmath: I had missed the link. Yes, this is essentially the same argument. Maybe the less formal explanation will help. – Ross Millikan May 10 '21 at 02:30
  • @hardmath: I added my understanding of the compactness argument. – Ross Millikan May 10 '21 at 02:51
  • But those axioms say nothing about the infinite subsets of $\mathbb N$, right? – hardmath May 10 '21 at 02:54
  • @hardmath: that is correct. I think it still works, which is the miracle of compactness. – Ross Millikan May 10 '21 at 02:57
  • Your compactness argument doesn't make any sense to me. How precisely are you expressing "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" as an axiom? – Eric Wofsey May 10 '21 at 03:02
  • @hardmath: I am thinking of the usual construction of nonstandard models of PA. You make a new element $c$ and add $c \gt n$ for all $n$. The usual naturals satisfy any finite set of these axioms but do not satisfy them all. – Ross Millikan May 10 '21 at 03:05