This is part of a bigger problem I'm trying to prove, but my argument relies of the validity of the following idea. Note that when I say countable, I don't mean finite — I mean countable infinity.
Consider the set of natural numbers, $\mathbb{N}$. Now take a countable subset of them, call it $\mathsf{S}_1$. Then take a countable subset of $\mathbb{N}\setminus{\mathsf{S}_1}$ and call it $\mathsf{S}_2$. Then take a countable subset of $\mathbb{N}\setminus\{{\mathsf{S}_1}\cup{\mathsf{S}_2}\}$ and call it $\mathsf{S}_3$. Continue in this manner to construct $\mathsf{S}_i$'s such that ${\mathsf{S}_i}\cap{\mathsf{S}_j}=\emptyset$ for $i≠j$.
My question is this: is it necessarily the case that eventually there will be finitely many $\mathsf{S}_i$'s with $\mathbb{N}\setminus\bigcup_{i}{\mathsf{S}_i}={P}$ where $P$ is either empty or finite? My guess is yes, but I'm not how to prove it or how to find a counterexample.
Best Answer
No, this isn't the case. For example, if we take $S_i$ to be the set of numbers $n$ such that $2^i\mid n$ but $2^{i+1}\nmid n$, then all of $S_i$ are countably infinite and pairwise disjoint. Hence taking out these sets will never leave you with a finite or empty set.