Are there any vector spaces that cannot be given a norm that makes the vector space a complete metric space?
$begingroup$
And if so, how can one prove that there is no such norm? I suppose one can use the form of Baire Category Theorem which states that a complete metric space cannot be written as a countable union of nowhere dense subsets to show that the metric space defined by the given vector space and norm is not complete but I have no idea on how to generalise this idea to all possible norms.
general-topology metric-spaces normed-spaces baire-category
asked Dec 19 '18 at 18:46
2chromatic2chromatic
1829
$endgroup$
add a comment |
$begingroup$
And if so, how can one prove that there is no such norm? I suppose one can use the form of Baire Category Theorem which states that a complete metric space cannot be written as a countable union of nowhere dense subsets to show that the metric space defined by the given vector space and norm is not complete but I have no idea on how to generalise this idea to all possible norms.
general-topology metric-spaces normed-spaces baire-category
asked Dec 19 '18 at 18:46
2chromatic2chromatic
1829
$endgroup$
add a comment |
$begingroup$
And if so, how can one prove that there is no such norm? I suppose one can use the form of Baire Category Theorem which states that a complete metric space cannot be written as a countable union of nowhere dense subsets to show that the metric space defined by the given vector space and norm is not complete but I have no idea on how to generalise this idea to all possible norms.
general-topology metric-spaces normed-spaces baire-category
asked Dec 19 '18 at 18:46
2chromatic2chromatic
1829
$endgroup$
And if so, how can one prove that there is no such norm? I suppose one can use the form of Baire Category Theorem which states that a complete metric space cannot be written as a countable union of nowhere dense subsets to show that the metric space defined by the given vector space and norm is not complete but I have no idea on how to generalise this idea to all possible norms.
general-topology metric-spaces normed-spaces baire-category
general-topology metric-spaces normed-spaces baire-category
asked Dec 19 '18 at 18:46
2chromatic2chromatic
1829
asked Dec 19 '18 at 18:46
2chromatic2chromatic
1829
asked Dec 19 '18 at 18:46
2chromatic2chromatic
1829
asked Dec 19 '18 at 18:46
2chromatic2chromatic
1829
asked Dec 19 '18 at 18:46
2chromatic2chromatic
1829
1829
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
If a space admits an enumerable non finite basis $(e_i)_{iin bf N}$ it cannot be complete, as it is the enuramble union of finite dimensional vector spaces $V_n= vect (e_1,...e_n)$.
answered Dec 19 '18 at 18:50
ThomasThomas
4,152510
$endgroup$
$begingroup$
A "natural" example is $mathbb{K}[X]$, $mathbb{K=R,C}$
$endgroup$
– Max
Dec 19 '18 at 20:50
add a comment |
$begingroup$
Yes. Take the space $c_{00}$ of all sequences $(a_n)_{ninmathbb N}$ of real numbers such that $a_n=0$ if $ngg1$. For each $ninmathbb N$, let $e(n)$ be the sequence such that its $n$th is $1$, where as all other terms are equal to $0$. Then, for any norm $lVertcdotrVert$ on $c_{00}$, the series$$sum_{n=0}^inftyfrac{e(n)}{n^2bigllVert e(n)bigrrVert}$$is a Cauchy series of elements of $c_{00}$ which doesn't converge in $c_{00}$.
answered Dec 19 '18 at 18:57
José Carlos SantosJosé Carlos Santos
172k22132239
$endgroup$
add a comment |
$begingroup$
The vector space $sigma$ of all sequences in $mathbb{R}^omega$ with the property that ${n: x_n neq 0}$ is finite (usually taken in the subspace topology induced from the product (the $sigma$-product this is sometimes called)) is a standard example.
It's a countable union of finite-dimensional subspaces. In any norm topology on $sigma$ the finite-dimensional subspaces are closed and have empty interior, so in any such norm the space cannot be Baire (and so cannot be complete).
answered Dec 19 '18 at 22:05
Henno BrandsmaHenno Brandsma
115k349125
$endgroup$
add a comment |
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3046734%2fare-there-any-vector-spaces-that-cannot-be-given-a-norm-that-makes-the-vector-sp%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If a space admits an enumerable non finite basis $(e_i)_{iin bf N}$ it cannot be complete, as it is the enuramble union of finite dimensional vector spaces $V_n= vect (e_1,...e_n)$.
answered Dec 19 '18 at 18:50
ThomasThomas
4,152510
$endgroup$
$begingroup$
A "natural" example is $mathbb{K}[X]$, $mathbb{K=R,C}$
$endgroup$
– Max
Dec 19 '18 at 20:50
add a comment |
$begingroup$
If a space admits an enumerable non finite basis $(e_i)_{iin bf N}$ it cannot be complete, as it is the enuramble union of finite dimensional vector spaces $V_n= vect (e_1,...e_n)$.
answered Dec 19 '18 at 18:50
ThomasThomas
4,152510
$endgroup$
$begingroup$
A "natural" example is $mathbb{K}[X]$, $mathbb{K=R,C}$
$endgroup$
– Max
Dec 19 '18 at 20:50
add a comment |
$begingroup$
If a space admits an enumerable non finite basis $(e_i)_{iin bf N}$ it cannot be complete, as it is the enuramble union of finite dimensional vector spaces $V_n= vect (e_1,...e_n)$.
answered Dec 19 '18 at 18:50
ThomasThomas
4,152510
$endgroup$
If a space admits an enumerable non finite basis $(e_i)_{iin bf N}$ it cannot be complete, as it is the enuramble union of finite dimensional vector spaces $V_n= vect (e_1,...e_n)$.
answered Dec 19 '18 at 18:50
ThomasThomas
4,152510
answered Dec 19 '18 at 18:50
ThomasThomas
4,152510
answered Dec 19 '18 at 18:50
ThomasThomas
4,152510
answered Dec 19 '18 at 18:50
ThomasThomas
4,152510
4,152510
$begingroup$
A "natural" example is $mathbb{K}[X]$, $mathbb{K=R,C}$
$endgroup$
– Max
Dec 19 '18 at 20:50
add a comment |
$begingroup$
A "natural" example is $mathbb{K}[X]$, $mathbb{K=R,C}$
$endgroup$
– Max
Dec 19 '18 at 20:50
$begingroup$
A "natural" example is $mathbb{K}[X]$, $mathbb{K=R,C}$
$endgroup$
– Max
Dec 19 '18 at 20:50
$begingroup$
A "natural" example is $mathbb{K}[X]$, $mathbb{K=R,C}$
$endgroup$
– Max
Dec 19 '18 at 20:50
add a comment |
$begingroup$
Yes. Take the space $c_{00}$ of all sequences $(a_n)_{ninmathbb N}$ of real numbers such that $a_n=0$ if $ngg1$. For each $ninmathbb N$, let $e(n)$ be the sequence such that its $n$th is $1$, where as all other terms are equal to $0$. Then, for any norm $lVertcdotrVert$ on $c_{00}$, the series$$sum_{n=0}^inftyfrac{e(n)}{n^2bigllVert e(n)bigrrVert}$$is a Cauchy series of elements of $c_{00}$ which doesn't converge in $c_{00}$.
answered Dec 19 '18 at 18:57
José Carlos SantosJosé Carlos Santos
172k22132239
$endgroup$
add a comment |
$begingroup$
Yes. Take the space $c_{00}$ of all sequences $(a_n)_{ninmathbb N}$ of real numbers such that $a_n=0$ if $ngg1$. For each $ninmathbb N$, let $e(n)$ be the sequence such that its $n$th is $1$, where as all other terms are equal to $0$. Then, for any norm $lVertcdotrVert$ on $c_{00}$, the series$$sum_{n=0}^inftyfrac{e(n)}{n^2bigllVert e(n)bigrrVert}$$is a Cauchy series of elements of $c_{00}$ which doesn't converge in $c_{00}$.
answered Dec 19 '18 at 18:57
José Carlos SantosJosé Carlos Santos
172k22132239
$endgroup$
add a comment |
$begingroup$
Yes. Take the space $c_{00}$ of all sequences $(a_n)_{ninmathbb N}$ of real numbers such that $a_n=0$ if $ngg1$. For each $ninmathbb N$, let $e(n)$ be the sequence such that its $n$th is $1$, where as all other terms are equal to $0$. Then, for any norm $lVertcdotrVert$ on $c_{00}$, the series$$sum_{n=0}^inftyfrac{e(n)}{n^2bigllVert e(n)bigrrVert}$$is a Cauchy series of elements of $c_{00}$ which doesn't converge in $c_{00}$.
answered Dec 19 '18 at 18:57
José Carlos SantosJosé Carlos Santos
172k22132239
$endgroup$
Yes. Take the space $c_{00}$ of all sequences $(a_n)_{ninmathbb N}$ of real numbers such that $a_n=0$ if $ngg1$. For each $ninmathbb N$, let $e(n)$ be the sequence such that its $n$th is $1$, where as all other terms are equal to $0$. Then, for any norm $lVertcdotrVert$ on $c_{00}$, the series$$sum_{n=0}^inftyfrac{e(n)}{n^2bigllVert e(n)bigrrVert}$$is a Cauchy series of elements of $c_{00}$ which doesn't converge in $c_{00}$.
answered Dec 19 '18 at 18:57
José Carlos SantosJosé Carlos Santos
172k22132239
answered Dec 19 '18 at 18:57
José Carlos SantosJosé Carlos Santos
172k22132239
answered Dec 19 '18 at 18:57
José Carlos SantosJosé Carlos Santos
172k22132239
answered Dec 19 '18 at 18:57
José Carlos SantosJosé Carlos Santos
172k22132239
172k22132239
add a comment |
add a comment |
$begingroup$
The vector space $sigma$ of all sequences in $mathbb{R}^omega$ with the property that ${n: x_n neq 0}$ is finite (usually taken in the subspace topology induced from the product (the $sigma$-product this is sometimes called)) is a standard example.
It's a countable union of finite-dimensional subspaces. In any norm topology on $sigma$ the finite-dimensional subspaces are closed and have empty interior, so in any such norm the space cannot be Baire (and so cannot be complete).
answered Dec 19 '18 at 22:05
Henno BrandsmaHenno Brandsma
115k349125
$endgroup$
add a comment |
$begingroup$
The vector space $sigma$ of all sequences in $mathbb{R}^omega$ with the property that ${n: x_n neq 0}$ is finite (usually taken in the subspace topology induced from the product (the $sigma$-product this is sometimes called)) is a standard example.
It's a countable union of finite-dimensional subspaces. In any norm topology on $sigma$ the finite-dimensional subspaces are closed and have empty interior, so in any such norm the space cannot be Baire (and so cannot be complete).
answered Dec 19 '18 at 22:05
Henno BrandsmaHenno Brandsma
115k349125
$endgroup$
add a comment |
$begingroup$
The vector space $sigma$ of all sequences in $mathbb{R}^omega$ with the property that ${n: x_n neq 0}$ is finite (usually taken in the subspace topology induced from the product (the $sigma$-product this is sometimes called)) is a standard example.
It's a countable union of finite-dimensional subspaces. In any norm topology on $sigma$ the finite-dimensional subspaces are closed and have empty interior, so in any such norm the space cannot be Baire (and so cannot be complete).
answered Dec 19 '18 at 22:05
Henno BrandsmaHenno Brandsma
115k349125
$endgroup$
The vector space $sigma$ of all sequences in $mathbb{R}^omega$ with the property that ${n: x_n neq 0}$ is finite (usually taken in the subspace topology induced from the product (the $sigma$-product this is sometimes called)) is a standard example.
It's a countable union of finite-dimensional subspaces. In any norm topology on $sigma$ the finite-dimensional subspaces are closed and have empty interior, so in any such norm the space cannot be Baire (and so cannot be complete).
answered Dec 19 '18 at 22:05
Henno BrandsmaHenno Brandsma
115k349125
answered Dec 19 '18 at 22:05
Henno BrandsmaHenno Brandsma
115k349125
answered Dec 19 '18 at 22:05
Henno BrandsmaHenno Brandsma
115k349125
answered Dec 19 '18 at 22:05
Henno BrandsmaHenno Brandsma
115k349125
115k349125
add a comment |
add a comment |
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3046734%2fare-there-any-vector-spaces-that-cannot-be-given-a-norm-that-makes-the-vector-sp%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
