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Given the equivalence relation: $A \sim B $ $\iff$ $A \Delta B$ is finite, where $\Delta$ is the symmetric difference, we define: $S := \{ A \subseteq \mathbb{N} \mid A \sim \mathbb{N} \}$. I am asked to calculate $\left| S \right|$.

Any tips, or suggestions?

Thank you!

Asaf Karagila
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Daniel
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    Sets in what you have labeled as $S$ here are known by a specific name... "Co-finite" sets. That is, they are the sets such that $\Bbb N\setminus A$ is finite. Make sure you can understand why these definitions match. – JMoravitz Jun 30 '23 at 14:04
  • With regards to the size of $S$... in the same way we can list finite sets: ${},{0},{1},{0,1},{2},{0,2},{1,2},{0,1,2},{3},{0,3},{1,3},\dots$ we can go about listing the co-finite sets in the same way... $\Bbb N\setminus {}, \Bbb N\setminus{0},\Bbb N\setminus {1},\dots$ – JMoravitz Jun 30 '23 at 14:06
  • I understand why the definitions match (because of how we define symmetric difference: $(A\setminus B) \cup (B \setminus A)$ ). What I'm still missing is the map, or rather how to define the set B such that $\mathbb{N} \setminus B$ is finite. I have a mental picture of taking $\mathbb{N}$ from the "end" and stopping each time leaving just a singleton, then a set containing 2 elements etc. but I'm not sure how to construct the proof of the lower bound. – Daniel Jun 30 '23 at 14:31
  • $\mathbb N\setminus A$ is finite iff ${k\geqslant N} \subseteq A$ for sufficiently large $N$, meaning there can't be infinitely many "holes" in $\mathbb N$ – AlvinL Jul 01 '23 at 07:06

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