Existence of a Banach space of arbitrary cardinal number $alphageq card( Bbb R)$












4












$begingroup$


Let $alpha$ be a cardinal number with $alphageq c:= operatorname{card}(Bbb R).$
Is there a Banach space $X$ which satisfies $operatorname{card}(X)= alpha?$
With many thanks for your answers.






set-theory banach-spaces







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$endgroup$








  • 5




    $begingroup$
    Two people downvoted this well formulated and polite question: that was a really stupid and mean move, especially after the system's warning "Be nice" to this new contributor.
    $endgroup$
    – Georges Elencwajg
    Jan 2 at 11:09








  • 1




    $begingroup$
    Yes. Consider the space $X$ of maps $sigma:alpha rightarrow mathbb{ R}$ with countable support, which are square summable. You can equip $X$ with a sufficiently complete norm as follows $Vert sigma Vert = Sigma { sigma(gamma)^2: gamma in supp(sigma)}$. ( $X$ is a linear space when equiped the pointwise addition and scalar mult.)
    $endgroup$
    – Not Mike
    Jan 2 at 11:20












  • $begingroup$
    I just want to know if this problem is open or it solve previously?
    $endgroup$
    – Ali Bayati
    Jan 2 at 11:23










  • $begingroup$
    If I had to guess, I'd guess this was solved close to a 100 years ago.
    $endgroup$
    – Asaf Karagila♦
    Jan 2 at 11:24










  • $begingroup$
    Are there any references in this area? Why we have $card( l^2(sigma))=card (sigma)$ when $card( sigma) geq c?$
    $endgroup$
    – Ali Bayati
    Jan 2 at 11:27


















4












$begingroup$


Let $alpha$ be a cardinal number with $alphageq c:= operatorname{card}(Bbb R).$
Is there a Banach space $X$ which satisfies $operatorname{card}(X)= alpha?$
With many thanks for your answers.






set-theory banach-spaces







share|cite|improve this question











$endgroup$








  • 5




    $begingroup$
    Two people downvoted this well formulated and polite question: that was a really stupid and mean move, especially after the system's warning "Be nice" to this new contributor.
    $endgroup$
    – Georges Elencwajg
    Jan 2 at 11:09








  • 1




    $begingroup$
    Yes. Consider the space $X$ of maps $sigma:alpha rightarrow mathbb{ R}$ with countable support, which are square summable. You can equip $X$ with a sufficiently complete norm as follows $Vert sigma Vert = Sigma { sigma(gamma)^2: gamma in supp(sigma)}$. ( $X$ is a linear space when equiped the pointwise addition and scalar mult.)
    $endgroup$
    – Not Mike
    Jan 2 at 11:20












  • $begingroup$
    I just want to know if this problem is open or it solve previously?
    $endgroup$
    – Ali Bayati
    Jan 2 at 11:23










  • $begingroup$
    If I had to guess, I'd guess this was solved close to a 100 years ago.
    $endgroup$
    – Asaf Karagila♦
    Jan 2 at 11:24










  • $begingroup$
    Are there any references in this area? Why we have $card( l^2(sigma))=card (sigma)$ when $card( sigma) geq c?$
    $endgroup$
    – Ali Bayati
    Jan 2 at 11:27
















4












4








4


1



$begingroup$


Let $alpha$ be a cardinal number with $alphageq c:= operatorname{card}(Bbb R).$
Is there a Banach space $X$ which satisfies $operatorname{card}(X)= alpha?$
With many thanks for your answers.






set-theory banach-spaces







share|cite|improve this question











$endgroup$




Let $alpha$ be a cardinal number with $alphageq c:= operatorname{card}(Bbb R).$
Is there a Banach space $X$ which satisfies $operatorname{card}(X)= alpha?$
With many thanks for your answers.






set-theory banach-spaces




set-theory banach-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 3 at 23:01









Davide Giraudo

127k16153268




127k16153268










asked Jan 2 at 11:00









Ali BayatiAli Bayati

344




344








  • 5




    $begingroup$
    Two people downvoted this well formulated and polite question: that was a really stupid and mean move, especially after the system's warning "Be nice" to this new contributor.
    $endgroup$
    – Georges Elencwajg
    Jan 2 at 11:09








  • 1




    $begingroup$
    Yes. Consider the space $X$ of maps $sigma:alpha rightarrow mathbb{ R}$ with countable support, which are square summable. You can equip $X$ with a sufficiently complete norm as follows $Vert sigma Vert = Sigma { sigma(gamma)^2: gamma in supp(sigma)}$. ( $X$ is a linear space when equiped the pointwise addition and scalar mult.)
    $endgroup$
    – Not Mike
    Jan 2 at 11:20












  • $begingroup$
    I just want to know if this problem is open or it solve previously?
    $endgroup$
    – Ali Bayati
    Jan 2 at 11:23










  • $begingroup$
    If I had to guess, I'd guess this was solved close to a 100 years ago.
    $endgroup$
    – Asaf Karagila♦
    Jan 2 at 11:24










  • $begingroup$
    Are there any references in this area? Why we have $card( l^2(sigma))=card (sigma)$ when $card( sigma) geq c?$
    $endgroup$
    – Ali Bayati
    Jan 2 at 11:27
















  • 5




    $begingroup$
    Two people downvoted this well formulated and polite question: that was a really stupid and mean move, especially after the system's warning "Be nice" to this new contributor.
    $endgroup$
    – Georges Elencwajg
    Jan 2 at 11:09








  • 1




    $begingroup$
    Yes. Consider the space $X$ of maps $sigma:alpha rightarrow mathbb{ R}$ with countable support, which are square summable. You can equip $X$ with a sufficiently complete norm as follows $Vert sigma Vert = Sigma { sigma(gamma)^2: gamma in supp(sigma)}$. ( $X$ is a linear space when equiped the pointwise addition and scalar mult.)
    $endgroup$
    – Not Mike
    Jan 2 at 11:20












  • $begingroup$
    I just want to know if this problem is open or it solve previously?
    $endgroup$
    – Ali Bayati
    Jan 2 at 11:23










  • $begingroup$
    If I had to guess, I'd guess this was solved close to a 100 years ago.
    $endgroup$
    – Asaf Karagila♦
    Jan 2 at 11:24










  • $begingroup$
    Are there any references in this area? Why we have $card( l^2(sigma))=card (sigma)$ when $card( sigma) geq c?$
    $endgroup$
    – Ali Bayati
    Jan 2 at 11:27










5




5




$begingroup$
Two people downvoted this well formulated and polite question: that was a really stupid and mean move, especially after the system's warning "Be nice" to this new contributor.
$endgroup$
– Georges Elencwajg
Jan 2 at 11:09






$begingroup$
Two people downvoted this well formulated and polite question: that was a really stupid and mean move, especially after the system's warning "Be nice" to this new contributor.
$endgroup$
– Georges Elencwajg
Jan 2 at 11:09






1




1




$begingroup$
Yes. Consider the space $X$ of maps $sigma:alpha rightarrow mathbb{ R}$ with countable support, which are square summable. You can equip $X$ with a sufficiently complete norm as follows $Vert sigma Vert = Sigma { sigma(gamma)^2: gamma in supp(sigma)}$. ( $X$ is a linear space when equiped the pointwise addition and scalar mult.)
$endgroup$
– Not Mike
Jan 2 at 11:20






$begingroup$
Yes. Consider the space $X$ of maps $sigma:alpha rightarrow mathbb{ R}$ with countable support, which are square summable. You can equip $X$ with a sufficiently complete norm as follows $Vert sigma Vert = Sigma { sigma(gamma)^2: gamma in supp(sigma)}$. ( $X$ is a linear space when equiped the pointwise addition and scalar mult.)
$endgroup$
– Not Mike
Jan 2 at 11:20














$begingroup$
I just want to know if this problem is open or it solve previously?
$endgroup$
– Ali Bayati
Jan 2 at 11:23




$begingroup$
I just want to know if this problem is open or it solve previously?
$endgroup$
– Ali Bayati
Jan 2 at 11:23












$begingroup$
If I had to guess, I'd guess this was solved close to a 100 years ago.
$endgroup$
– Asaf Karagila♦
Jan 2 at 11:24




$begingroup$
If I had to guess, I'd guess this was solved close to a 100 years ago.
$endgroup$
– Asaf Karagila♦
Jan 2 at 11:24












$begingroup$
Are there any references in this area? Why we have $card( l^2(sigma))=card (sigma)$ when $card( sigma) geq c?$
$endgroup$
– Ali Bayati
Jan 2 at 11:27






$begingroup$
Are there any references in this area? Why we have $card( l^2(sigma))=card (sigma)$ when $card( sigma) geq c?$
$endgroup$
– Ali Bayati
Jan 2 at 11:27












1 Answer
1






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oldest

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6












$begingroup$

The answer is no.



Suppose that $alpha=beth_omega$, where $beth_0=aleph_0$ and $beth_{alpha+1}=2^{beth_alpha}$, with supremum at limits. Then $alpha>frak c$.



If $X$ is a Banach space of cardinality $alpha$, its dimension over $Bbb R$ is $alpha$ as well, fix a basis of size $alpha$ and let $X_n$ be the closed span of the first $beth_n$ vectors in the basis. Since taking closure uses only countable sequences, and $$beth_n^{aleph_0}leqbeth_n^{beth_n}=2^{beth_n}=beth_{n+1}$$
it follows that $X_n$ is a closed subspace of size at most $beth_{n+1}$. In particular, it has empty interior, since any open ball must have size $|X|$. But it is also easy to see that $bigcup X_n=X$, which contradicts the Baire Category Theorem.



This can be generalized to all $alpha$ such that $alpha^{aleph_0}>alpha$. And indeed that is the only limitation. If $|S|=alpha$ such that $alpha^{aleph_0}=alpha$, then $ell^infty(S)$ has size $alpha$ and a natural Banach space structure.






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    6












    $begingroup$

    The answer is no.



    Suppose that $alpha=beth_omega$, where $beth_0=aleph_0$ and $beth_{alpha+1}=2^{beth_alpha}$, with supremum at limits. Then $alpha>frak c$.



    If $X$ is a Banach space of cardinality $alpha$, its dimension over $Bbb R$ is $alpha$ as well, fix a basis of size $alpha$ and let $X_n$ be the closed span of the first $beth_n$ vectors in the basis. Since taking closure uses only countable sequences, and $$beth_n^{aleph_0}leqbeth_n^{beth_n}=2^{beth_n}=beth_{n+1}$$
    it follows that $X_n$ is a closed subspace of size at most $beth_{n+1}$. In particular, it has empty interior, since any open ball must have size $|X|$. But it is also easy to see that $bigcup X_n=X$, which contradicts the Baire Category Theorem.



    This can be generalized to all $alpha$ such that $alpha^{aleph_0}>alpha$. And indeed that is the only limitation. If $|S|=alpha$ such that $alpha^{aleph_0}=alpha$, then $ell^infty(S)$ has size $alpha$ and a natural Banach space structure.






    share|cite|improve this answer











    $endgroup$


















      6












      $begingroup$

      The answer is no.



      Suppose that $alpha=beth_omega$, where $beth_0=aleph_0$ and $beth_{alpha+1}=2^{beth_alpha}$, with supremum at limits. Then $alpha>frak c$.



      If $X$ is a Banach space of cardinality $alpha$, its dimension over $Bbb R$ is $alpha$ as well, fix a basis of size $alpha$ and let $X_n$ be the closed span of the first $beth_n$ vectors in the basis. Since taking closure uses only countable sequences, and $$beth_n^{aleph_0}leqbeth_n^{beth_n}=2^{beth_n}=beth_{n+1}$$
      it follows that $X_n$ is a closed subspace of size at most $beth_{n+1}$. In particular, it has empty interior, since any open ball must have size $|X|$. But it is also easy to see that $bigcup X_n=X$, which contradicts the Baire Category Theorem.



      This can be generalized to all $alpha$ such that $alpha^{aleph_0}>alpha$. And indeed that is the only limitation. If $|S|=alpha$ such that $alpha^{aleph_0}=alpha$, then $ell^infty(S)$ has size $alpha$ and a natural Banach space structure.






      share|cite|improve this answer











      $endgroup$
















        6












        6








        6





        $begingroup$

        The answer is no.



        Suppose that $alpha=beth_omega$, where $beth_0=aleph_0$ and $beth_{alpha+1}=2^{beth_alpha}$, with supremum at limits. Then $alpha>frak c$.



        If $X$ is a Banach space of cardinality $alpha$, its dimension over $Bbb R$ is $alpha$ as well, fix a basis of size $alpha$ and let $X_n$ be the closed span of the first $beth_n$ vectors in the basis. Since taking closure uses only countable sequences, and $$beth_n^{aleph_0}leqbeth_n^{beth_n}=2^{beth_n}=beth_{n+1}$$
        it follows that $X_n$ is a closed subspace of size at most $beth_{n+1}$. In particular, it has empty interior, since any open ball must have size $|X|$. But it is also easy to see that $bigcup X_n=X$, which contradicts the Baire Category Theorem.



        This can be generalized to all $alpha$ such that $alpha^{aleph_0}>alpha$. And indeed that is the only limitation. If $|S|=alpha$ such that $alpha^{aleph_0}=alpha$, then $ell^infty(S)$ has size $alpha$ and a natural Banach space structure.






        share|cite|improve this answer











        $endgroup$



        The answer is no.



        Suppose that $alpha=beth_omega$, where $beth_0=aleph_0$ and $beth_{alpha+1}=2^{beth_alpha}$, with supremum at limits. Then $alpha>frak c$.



        If $X$ is a Banach space of cardinality $alpha$, its dimension over $Bbb R$ is $alpha$ as well, fix a basis of size $alpha$ and let $X_n$ be the closed span of the first $beth_n$ vectors in the basis. Since taking closure uses only countable sequences, and $$beth_n^{aleph_0}leqbeth_n^{beth_n}=2^{beth_n}=beth_{n+1}$$
        it follows that $X_n$ is a closed subspace of size at most $beth_{n+1}$. In particular, it has empty interior, since any open ball must have size $|X|$. But it is also easy to see that $bigcup X_n=X$, which contradicts the Baire Category Theorem.



        This can be generalized to all $alpha$ such that $alpha^{aleph_0}>alpha$. And indeed that is the only limitation. If $|S|=alpha$ such that $alpha^{aleph_0}=alpha$, then $ell^infty(S)$ has size $alpha$ and a natural Banach space structure.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 2 at 14:43









        David C. Ullrich

        61.2k43994




        61.2k43994










        answered Jan 2 at 11:23









        Asaf Karagila♦Asaf Karagila

        305k33436767




        305k33436767






























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