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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 )

  • Why are the questions posed in title and body different? – k.stm Feb 16 '16 at 13:46
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    Yes. No. ${}{}{}{}{}{}{}{}{}$ – David C. Ullrich Feb 16 '16 at 13:46
  • @To all : I think things are ok now –  Feb 16 '16 at 13:50
  • Hint: Two vector spaces of equal dimension are isomorphic as vector spaces (over a given field). – hardmath Feb 16 '16 at 13:51
  • @SaunDev I took the liberty of changing the title into a somewhat more concise question (for the sake of readability). Oh yeah, forgot uncountable. – k.stm Feb 16 '16 at 13:52
  • @k.stm : But uncountable dimensional is necessary otherwise the answer would be trivially no –  Feb 16 '16 at 13:52
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    @hardmath : Ok , so you are telling me to look , for each uncountable cardinal , to a real vector space which is complete ? Or something else ? –  Feb 16 '16 at 13:54
  • @SaunDev: Yes, at least if I understood the question. I did have some doubt whether you start with a topological vector space of uncountable dimension and expect to realize it by a complete norm (or even metric), which is clearly untenable. But if you only have to preserve the dimension of $V$ as a vector space, then that's the approach I'd take. – hardmath Feb 16 '16 at 13:57
  • @hardmath : Well I just need to preserve the dimension and algebraic properties , I am assuming no topological properties of the space beforehand . So what is(are) the familiar space(s) you are talking about ? Nothing really comes to mind :( –  Feb 16 '16 at 14:01
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    No, if the Hamel dimension is $\aleph_\omega$, then it is not a Banach space. – GEdgar Feb 16 '16 at 14:14
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    Strongly related – Daniel Fischer Feb 16 '16 at 14:17
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    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. – Daniel Fischer Feb 16 '16 at 14:20
  • @DanielFischer : Ok , so there is some very involved set theoretic concepts related too , I see ... –  Feb 16 '16 at 14:24

1 Answers1

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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 $V$.

Christian Blatter
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    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) $\alpha$ " ? –  Feb 16 '16 at 14:09