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$A$ is an infinite set.

Let $F$ be the set of all finite subsets of $A$, prove that $\operatorname{card}(F)=\operatorname{card}(A)$.

I want to prove this conclusion:

$\operatorname{card}\left(F\right)\leq \operatorname{card}\left(\bigsqcup_n \mathcal{F}\left(\left[n\right],A\right)\right)$

where $\left[n\right]$ is a n-tuple set and $\mathcal{F}\left(X,Y\right)$ is a mapping of $X$ onto $Y$.

Since after that $\text{LHS}\geq \operatorname{card}\left(A\right)$

$\text{RHS}\leq \operatorname{card}\left(\mathbb{N} \times A\right)\leq \operatorname{card}\left(A\times A\right)\leq \operatorname{card}\left(A\right)$.

J. W. Tanner
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Quyi Le
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  • Why is "$\text{RHS}\leq \operatorname{card}\left(\mathbb{N} \times A\right)$"? I'd rather begin with $\text{RHS}\le\operatorname{card}(\cup_{n\in\Bbb N}A^n).$ – Anne Bauval Feb 09 '23 at 18:08
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    Yes, I start with that one and I already prove that if S is a finite set, then $card(\mathcal{F}(S,A))=card(A)$ so I miss it. – Quyi Le Feb 10 '23 at 01:59

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