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On any real vector space $V$ of uncountable dimension , can we always define a norm such that endowed with that norm , $V$ becomes a complete normed linear space ? ( I know it can be done if $V$ is finite dimensional but what if $V$ is infinite dimensional ? The only thing I know is that any infinite dimensional Banach space must be of uncountable dimension )

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    $\begingroup$ Yes. No. ${}{}{}{}{}{}{}{}{}$ $\endgroup$ – David C. Ullrich Feb 16 '16 at 13:46
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    $\begingroup$ @hardmath : Ok , so you are telling me to look , for each uncountable cardinal , to a real vector space which is complete ? Or something else ? $\endgroup$ – user228168 Feb 16 '16 at 13:54
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    $\begingroup$ No, if the Hamel dimension is $\aleph_\omega$, then it is not a Banach space. $\endgroup$ – GEdgar Feb 16 '16 at 14:14
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    $\begingroup$ Strongly related $\endgroup$ – Daniel Fischer Feb 16 '16 at 14:17
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    $\begingroup$ Apparently, with the axiom of choice, a necessary and sufficient condition for a Banach space of infinite (algebraic) dimension $\kappa$ to exist is that $\kappa^{\aleph_0} = \kappa$. If there is an infinite Dedekind-finite set, however, that condition is no longer necessary. $\endgroup$ – Daniel Fischer Feb 16 '16 at 14:20
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I'm not a set theory expert, but I'd argue as follows:

Begin with your favorite Banach space $B$, forget its metric, and consider only its vector space structure. As a vector space, $B$ possesses a Hamel base $(e_\iota)_{\iota\in I}$ of a certain cardinality $|I|$.

Now if the dimension of your vector space $V$ equals $|I|$ you can copy the Banach space structure of $B$ over to $V$.

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    $\begingroup$ Yes , yes that is true , then the question reduces to " Does for every uncountable cardinal $\alpha$ , there exists a Banach space of dimension (Hamel basis dimension) $\alpha$ " ? $\endgroup$ – user228168 Feb 16 '16 at 14:09

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