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Show that the set of all finite subsets of $\mathbb{N}$ is countable.

How can I prove in a proper way that the "set of all finite subsets of $\mathbb{N}$ (the set of natural numbers) is a countable set"? Please help me with this.Thank you.

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Grazel: I think you'll find some nice answers to your question by clicking on the link immediately above. –  amWhy Dec 22 '12 at 15:24
    
Also see: math.stackexchange.com/questions/263677/… –  Asaf Karagila Dec 22 '12 at 15:27
    
Since $\mathbb{N}$ is a countably infinite set, it may help to see as well math.stackexchange.com/questions/27096/…. –  LePressentiment Nov 7 '13 at 1:27
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marked as duplicate by amWhy, Asaf Karagila, Davide Giraudo, MJD, Nameless Dec 22 '12 at 16:34

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2 Answers

Let $S$ be the set of all finite subsets of $N$. It is easy to verify that the function $f:S\rightarrow N$ that sends $A$ to $\sum_{a\in A} 2^{a}$ is a bijection

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For an approach with cardinal numbers:

  • The set of all finite subsets of $\mathbb{N}$ is the union over $n$ of the set of all subsets of size $n$.
  • The set of all subsets of size $n$ is no bigger than the set of all ordered tuples of length $n$.

Therefore, the set of all finite subsets of $\mathbb{N}$ has cardinality less than or equal to:

$$ \sum_{i \in \mathbb{N}} |\mathbb{N}|^i = \sum_{i \in \mathbb{N}} |\mathbb{N}| = |\mathbb{N}| \cdot |\mathbb{N}| = |\mathbb{N}|$$

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