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I am trying to find the cardinality of the set $A = \{E \subseteq \mathbb{N}: \exists p \text{ prime},m \in \mathbb{N} \text{ such that }|E| = p^m\}.$

Since the set of all primes has cardinality $\aleph_0$ I feel like there could be infinitely many sets in $\mathbb{N}$ with prime cardinality. So $|A| \geq \aleph_0$. Is my intuition correct?

Anne Bauval
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Saim Faigol
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    This is a strict subset of the set of finite subsets of $\mathbb{N}$, which is countable. It's also infinite, as you note, so your set has size $\aleph_0$. – Connor W Jul 09 '23 at 03:30
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    @Taladris, yes, but each set in $A$ is of finite size. This means that $A\subseteq F(\mathbb{N}),$ where $F(\mathbb{N})$ is the set of all finite subsets of $\mathbb{N}.$ In fact, $A\subset F(\mathbb{N}),$ as ${2}\in F(\mathbb{N})$ but ${2}\notin A.$ – aqualubix Jul 09 '23 at 04:14
  • Please edit your post to clarify your question: is it your title or only your final sentence? – Anne Bauval Jul 09 '23 at 06:22
  • Instead of editing an answer, we should find some duplicate of @ConnorW's comment. – Anne Bauval Jul 09 '23 at 06:24
  • So, I should be able to find an injection $f: A \to F(\mathbb{N})$, right? @Connor W – Saim Faigol Jul 09 '23 at 13:25
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    You have to find an injection from $A$ to $\mathbb{N}$. Since $A$ is already a subset of $F(\mathbb{N})$, so there is an obvious injection. As $F(N)$ is countable, it has an injection into $\mathbb{N}$. Now compose these two. @SaimFaigol – Kalas678 Jul 09 '23 at 13:33
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    See for example this frequently duplicated Question, Show that the set of all finite subsets of $\mathbb N$ is countable. I'm guessing that with that insight the road to an Answer is clear to the Original Poster. A contextual remark about why the problem statement here is motivated would be welcome. – hardmath Jul 09 '23 at 13:57

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