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I found the demonstration of this assertion in a book:

"Let the cardinality of some Hamel basis be $\kappa$. We can easily calculate the cardinality of the generated vector space: it is $\aleph_0(\kappa+\kappa^2+\ldots)=\aleph_0\kappa = \kappa$ and since this must be equal to c, we obtain $\kappa$ = c ." (c is the cardinality of the continuum)

How did they get $\aleph_0(\kappa+\kappa^2+\ldots)$ ?

creative
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ketherok
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    Basis of what? ${}{}$ – user251257 Apr 04 '17 at 17:49
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    If this is a basis of the reals over the rationals; or something like this, the $\aleph_0$ is for the cardinality of the set of coefficients. one then gets $\aleph_0\kappa$ elements using each basis element like $qb$, $(\aleph_0 \kappa)^2$ elements using two basis elements $q_1b_1+ q_2 b_2$, and so on. Then use that $\aleph_0^2 $ is $\aleph_0$. – quid Apr 04 '17 at 17:52

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I think Quid's comment above is exactly to the point, so I am reposting it CW.

If this is a basis of the reals over the rationals; or something like this, the $ℵ_0$ is for the cardinality of the set of coefficients. one then gets $ℵ_0κ$ elements using each basis element like $qb$, $(ℵ_0κ)^2$ elements using two basis elements $q_1b_1+q_2b_2$, and so on. Then use that $ℵ_0^2$ is $ℵ_0$.

The span of the basis is then $$ \begin{align} & \hphantom{=} \aleph_0k + \aleph_0^2k^2 + \aleph_0^3k^3 + \ldots\\ & = \aleph_0k + \aleph_0k^2 + \aleph_0k^3 + \ldots \\ & = \color{darkred}{\aleph_0(k +k^2 + k^3 + \ldots) } \\ & = \aleph_0k\\ & = k \end{align} $$

MJD
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  • it is effectively a basis of the reals over the rationals, thank you. How do you get the two last equalities ? For example for the last one, I thought that the product of cardinal $\kappa\cdot\mu$ one of which is infinite (let's say $\mu$) is equal to $\mu$. Here $\aleph_0$ is infinite but we dont't know about $k$. – ketherok Apr 04 '17 at 22:33
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    If $k$ were finite or countably infinite, its span would be countably infinite. But we know it spans the reals, which are not finite or countably infinite. So $k$ is certainly uncountable. The product of two infinite cardinals is equal to whichever is bigger. So $\aleph_0k = k$. – MJD Apr 04 '17 at 23:43
  • ah ok since $\aleph_0$ is the smallest infinite cardinal the product must be equal to $k$, thank you. – ketherok Apr 05 '17 at 07:16
  • Thanks for the repost and the addition. – quid Apr 06 '17 at 23:28