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$\{S \mid S ⊆ \Bbb N, |S| = \infty\}$

I would have thought that the set would be finite since S belongs to the natural numbers which are countable. Wouldn't the set above be a subset of a countable set which would also make it countable?

Mauro ALLEGRANZA
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FarmerZee
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    The set is a collection of subsets of $\mathbb{N}$. Hence, it is actually a subset of the power set $\mathcal{P} \left( \mathbb{N} \right)$, which is uncountable. That is, the given set need not be countable. – Aniruddha Deshmukh Oct 28 '21 at 05:26
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    This set is uncountable since the collection ${S\ |\ S \subseteq \mathbb N, |S|\ \text {is finite} }$ is countable and $|\mathcal P (\mathbb N)| = 2^{\aleph_{0}} = \mathfrak c.$ For otherwise $\mathcal P (\mathbb N)$ can be written as union of two countable sets and hence it would itself become countable, a contradiction. – Anil Bagchi. Oct 28 '21 at 05:47
  • @AntonioClaire Why not an official answer? – Paul Frost Oct 28 '21 at 09:04
  • Your set is definitely not finite. It contains all $S_n = \mathbb N \setminus {n}$ with $n \in \mathbb N$. – Paul Frost Oct 28 '21 at 09:05
  • @Paul Frost$:$ Actually I was busy with some of my works. I guessed that OP could be able to figure it out on his/her own after reading the first two comments; first one by Aniruddha followed by the one by me. – Anil Bagchi. Oct 29 '21 at 05:14
  • @AntonioClaire Yes, of course. My comment only concerned the sentence "I would have thought that the set would be finite" in the question. – Paul Frost Oct 29 '21 at 08:24

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Let \begin{align*} \mathcal{P}(\Bbb N)&=\{S \mid S ⊆ \Bbb N\},\\ [\Bbb N]^\omega&=\{S \mid S ⊆ \Bbb N, |S| = \omega\}=\{S \mid S ⊆ \Bbb N, S\text{ is infinite}\},\\ [\Bbb N]^{<\omega}&=\{S \mid S ⊆ \Bbb N, |S| < \omega\}=\{S \mid S ⊆ \Bbb N, S\text{ is finite}\}.\\ \end{align*}

Then

$$\mathcal{P}(\Bbb N)=[\Bbb N]^{\omega}\cup [\Bbb N]^{<\omega},$$

and so

$$2^{\aleph_0}=|[\Bbb N]^{\omega}|+{\aleph_0}$$

which follows that

$$|[\Bbb N]^{\omega}|=2^{\aleph_0}.$$

Furthermore, $|[A]^{\omega}|=2^{\aleph_0}$ for any countable set $A$.

M. Logic
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