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







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












$begingroup$


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

127k16151265




127k16151265










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






active

oldest

votes


















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











$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3059345%2fexistence-of-a-banach-space-of-arbitrary-cardinal-number-alpha-geq-card-bbb%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    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

        61k43994




        61k43994










        answered Jan 2 at 11:23









        Asaf Karagila♦Asaf Karagila

        305k33435766




        305k33435766






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3059345%2fexistence-of-a-banach-space-of-arbitrary-cardinal-number-alpha-geq-card-bbb%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Ellipse (mathématiques)

            Image
            Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
            Read more

            Quarter-circle Tiles

            Image
            up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
            Read more

            Mont Emei

            Image
            Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
            Read more