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As a starting point I have started to use the fact, that every subset of $\mathbb{N}$ countable is. A $\subset \mathbb{N}$ is also a subset of $ \mathbb{N}$ and at this point I can't go forward.

Or can I eventually use somehow Hilbert's Hotel theorem?

I appreciate any help a lot.

Herrpeter
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    Are you sure the title is correct? I think you meant the set of all finite subsets of $\mathbb{N}$ is countable. –  May 27 '21 at 21:01
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    “Should I show that it is surjective and injective?” You don’t have a function in sight; what could you possibly show is “surjective and injective”? – Arturo Magidin May 27 '21 at 21:02
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    It's important to note that "Hilbert's hotel" is not an actual mathematical argument. It's a popularized version of the fact that a function from the natural numbers to itself may be injective without being surjective. Hilbert mentioned it once in 1924, in a popular lecture, and never mentioned it again in his life. The story was picked up in a popular book by physicist George Gamow in 1947, and now has mindshare among religious charlatans (William Lane Craig) to make a point about infinite regress; and many people, philosophers and math students, who have been taught that ... (continued) – user4894 May 27 '21 at 21:22
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    (more) ... it's a mathematical argument. It is not. Interested readers would enjoy science historian Helge Kragh's definitive account of the history of the story. https://arxiv.org/abs/1403.0059 – user4894 May 27 '21 at 21:25

2 Answers2

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The question as stated is not true. I will prove what is true - the set $M=\{A\subset \Bbb{N}:A \ \text{is finite}\}$ is countable. We denote by $M_n$ all finite subsets of $\Bbb{N}$ s.t their cardinality is exactly $n$. What is the cardinality of each set? Well, I claim that $|M_n|=\aleph_0^n=\aleph_0$. Simply define a function $\phi:M_n\rightarrow \Bbb{N}^n$ by - for every $A_n\in M_n,\phi(A_n)$ is simply the vector of length $n$ with it's coordinates being exactly the element of $A_n$, ordered by the natural order of $\Bbb{N}$.

This is obviously an injective function - and so $|M_n|=\aleph_0^n=\aleph_0$. So, since $M=\bigcup_{n\in_\Bbb{N}}M_n$, we now see that $M$ is a countable union of countable sets and therefore countable.

The set you mentioned is ofcourse not countable, being exactly equal to $P(\Bbb{N})\setminus M$, and since $|P(\Bbb{N})|=\aleph$ and $|M|=\aleph_0$, we see that $|P(\Bbb{N})\setminus M|=\aleph$

Math101
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You cannot show that statement as it is false. As Micheal Barz pointed out in the comments you are probably looking for this statement here Show that the set of all finite subsets of $\mathbb{N}$ is countable..

We can exhibit an uncountable subset of $M$. Namely, we have $$ N:=\{ A \subseteq \mathbb{N} \ : \ 2\mathbb{N} +1\subseteq A \} \subseteq M. $$ However, we have the following bijections $$ f: \mathcal{P}(\mathbb{N}) \rightarrow \mathcal{P}(2\mathbb{N}), S \mapsto 2S $$ and $$g: \mathcal{P}(2\mathbb{N}) \rightarrow N, B \mapsto B \cup (2\mathbb{N}+1).$$ We know that $\mathcal{P}(\mathbb{N})$ is uncountable and thus, we have that $M$ and $N$ are uncountable too.

Severin Schraven
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  • Maybe what you mean is to construct a bijection $f : \mathcal{P}(\mathbb{N}) \to { A \subseteq \mathbb{N} \mid 2\mathbb{N} \subseteq A }$? In which case, you would need something more like $S \mapsto 2\mathbb{N} \cup (2S + 1)$ (this formula assuming $0 \in \mathbb{N}$ - if instead you prefer $\mathbb{N}$ starting at 1, the formula would become $S \mapsto 2\mathbb{N} \cup (2S - 1)$). – Daniel Schepler May 27 '21 at 21:26
  • @DanielSchepler Thanks for your comment. I actually wanted $2\mathbb{N}+1\subseteq A$, then this set is in bijection with $\mathcal{P}(2\mathbb{N}).$ – Severin Schraven May 28 '21 at 07:49