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I want to prove that if $A$ is a infinite set, then $|Fin(A)|=|FS(A)|=|A|$, where $Fin(A)$ is the set of all finite subsets of $A$ and $|FS(A)|$ is the set of all finite sequences. Firstly, to prove $|Fin(A)|=|A|$,
$$|Fin(A)|=|\bigcup_{n < \omega}[A]^{n}|=|\bigcup_{n < \omega}A|=\sum_{n <\omega}|A|=|A|\aleph_0=|A|$$ where in the second equality i use that $|[A]^{n}|=|A^{n}|=|A|$ (i'm also using that $|A \times A|=|A|$, thanks axiom of choice).
For the second, I would to use a similar argument and write $FS(A)=\bigcup_{n < \omega}A^{n}$, but i'm not truly sure about that equality.

user1868607
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avir_12
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  • Well, it's equivalent to choice. So I'm not sure what your question is here. – Asaf Karagila May 29 '20 at 10:57
  • Your equality for $FS(A)$ is correct. – Berci May 29 '20 at 11:11
  • The [axiom-of-choice] tag suggests that you're asking about the necessity of the axiom of choice here, which would fit well with the [set-theory] tag as @Andrés left it. But the content of the question seems more like a proof verification, which may be more appropriate to [elementary-set-theory] instead. So it would be very helpful if you can clarify your question. – Asaf Karagila May 29 '20 at 13:27
  • @Berci How could I prove the equality? – avir_12 May 29 '20 at 18:05
  • Well, what's your definition for finite sequence of $A$? – Berci May 29 '20 at 18:18
  • FS(A) is the set of all ordered finite sequences of element of A. – avir_12 May 29 '20 at 18:31

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It appears to me that you are asking about the case of finite sequences as you worked out the finite subsets yourself. There are also other questions that think about finite sets: The cardinality of the set of all finite subsets of an infinite set. Here is my proof of your theorem.

If $Card(A) = \kappa$ is an infinite cardinal then $Card(A \times A) = Card(A^{< \omega}) = \kappa$.

Proof

$Card(A \times A) = Card(A) \cdot Card(A) = \kappa \cdot \kappa = \max(\kappa,\kappa) = \kappa$.

Clearly, by finite induction, it follows that $Card(A^n) = \kappa$ for all $n < \omega$.

Now, observe that $A^{< \omega} = \bigcup_{n < \omega} A^n$ is bijective with $\mathbb{N} \times A$ or equivalently with $\mathbb{N} \times \kappa$. So: $Card(A^{< \omega}) = Card(\mathbb{N} \times \kappa) = \mathbb{N} \cdot A = \max(\mathbb{N}, \kappa) = \kappa$.

The bijection seems easy, choose $x \in \bigcup_{n < \omega} A^n$ then $x \in A^i$ for a unique $i$ (indeed, this union must be disjoint since each component has a different length). Then assign $x$ to $(i,x)$. The inverse would assign $(i,x)$ to the element $x$ of $A^i$.

user1868607
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