Existence of a Banach space of arbitrary cardinal number $alphageq card( Bbb R)$
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
edited Jan 3 at 23:01
Davide Giraudo
125k16150260
asked Jan 2 at 11:00
Ali BayatiAli Bayati
263
New contributor
Ali Bayati is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
|
show 2 more comments
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
edited Jan 3 at 23:01
Davide Giraudo
125k16150260
asked Jan 2 at 11:00
Ali BayatiAli Bayati
263
New contributor
Ali Bayati is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
4
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.
– Georges Elencwajg
Jan 2 at 11:09
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.)
– Not Mike
Jan 2 at 11:20
I just want to know if this problem is open or it solve previously?
– Ali Bayati
Jan 2 at 11:23
If I had to guess, I'd guess this was solved close to a 100 years ago.
– Asaf Karagila♦
Jan 2 at 11:24
Are there any references in this area? Why we have $card( l^2(sigma))=card (sigma)$ when $card( sigma) geq c?$
– Ali Bayati
Jan 2 at 11:27
|
show 2 more comments
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
edited Jan 3 at 23:01
Davide Giraudo
125k16150260
asked Jan 2 at 11:00
Ali BayatiAli Bayati
263
New contributor
Ali Bayati is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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
edited Jan 3 at 23:01
Davide Giraudo
125k16150260
asked Jan 2 at 11:00
Ali BayatiAli Bayati
263
New contributor
Ali Bayati is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Jan 3 at 23:01
Davide Giraudo
125k16150260
asked Jan 2 at 11:00
Ali BayatiAli Bayati
263
New contributor
Ali Bayati is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Jan 3 at 23:01
Davide Giraudo
125k16150260
edited Jan 3 at 23:01
Davide Giraudo
125k16150260
edited Jan 3 at 23:01
Davide Giraudo
125k16150260
125k16150260
asked Jan 2 at 11:00
Ali BayatiAli Bayati
263
New contributor
Ali Bayati is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 2 at 11:00
Ali BayatiAli Bayati
263
asked Jan 2 at 11:00
Ali BayatiAli Bayati
263
263
New contributor
Ali Bayati is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Ali Bayati is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Ali Bayati is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
4
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.
– Georges Elencwajg
Jan 2 at 11:09
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.)
– Not Mike
Jan 2 at 11:20
I just want to know if this problem is open or it solve previously?
– Ali Bayati
Jan 2 at 11:23
If I had to guess, I'd guess this was solved close to a 100 years ago.
– Asaf Karagila♦
Jan 2 at 11:24
Are there any references in this area? Why we have $card( l^2(sigma))=card (sigma)$ when $card( sigma) geq c?$
– Ali Bayati
Jan 2 at 11:27
|
show 2 more comments
4
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.
– Georges Elencwajg
Jan 2 at 11:09
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.)
– Not Mike
Jan 2 at 11:20
I just want to know if this problem is open or it solve previously?
– Ali Bayati
Jan 2 at 11:23
If I had to guess, I'd guess this was solved close to a 100 years ago.
– Asaf Karagila♦
Jan 2 at 11:24
Are there any references in this area? Why we have $card( l^2(sigma))=card (sigma)$ when $card( sigma) geq c?$
– Ali Bayati
Jan 2 at 11:27
4
4
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.
– Georges Elencwajg
Jan 2 at 11:09
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.
– Georges Elencwajg
Jan 2 at 11:09
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.)
– Not Mike
Jan 2 at 11:20
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.)
– Not Mike
Jan 2 at 11:20
I just want to know if this problem is open or it solve previously?
– Ali Bayati
Jan 2 at 11:23
I just want to know if this problem is open or it solve previously?
– Ali Bayati
Jan 2 at 11:23
If I had to guess, I'd guess this was solved close to a 100 years ago.
– Asaf Karagila♦
Jan 2 at 11:24
If I had to guess, I'd guess this was solved close to a 100 years ago.
– Asaf Karagila♦
Jan 2 at 11:24
Are there any references in this area? Why we have $card( l^2(sigma))=card (sigma)$ when $card( sigma) geq c?$
– Ali Bayati
Jan 2 at 11:27
Are there any references in this area? Why we have $card( l^2(sigma))=card (sigma)$ when $card( sigma) geq c?$
– Ali Bayati
Jan 2 at 11:27
|
show 2 more comments
1 Answer
1
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oldest
votes
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.
edited Jan 2 at 14:43
David C. Ullrich
59.1k43892
answered Jan 2 at 11:23
Asaf Karagila♦Asaf Karagila
302k32426757
add a comment |
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1 Answer
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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.
edited Jan 2 at 14:43
David C. Ullrich
59.1k43892
answered Jan 2 at 11:23
Asaf Karagila♦Asaf Karagila
302k32426757
add a comment |
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.
edited Jan 2 at 14:43
David C. Ullrich
59.1k43892
answered Jan 2 at 11:23
Asaf Karagila♦Asaf Karagila
302k32426757
add a comment |
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.
edited Jan 2 at 14:43
David C. Ullrich
59.1k43892
answered Jan 2 at 11:23
Asaf Karagila♦Asaf Karagila
302k32426757
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.
edited Jan 2 at 14:43
David C. Ullrich
59.1k43892
answered Jan 2 at 11:23
Asaf Karagila♦Asaf Karagila
302k32426757
edited Jan 2 at 14:43
David C. Ullrich
59.1k43892
edited Jan 2 at 14:43
David C. Ullrich
59.1k43892
edited Jan 2 at 14:43
David C. Ullrich
59.1k43892
59.1k43892
answered Jan 2 at 11:23
Asaf Karagila♦Asaf Karagila
302k32426757
answered Jan 2 at 11:23
Asaf Karagila♦Asaf Karagila
302k32426757
answered Jan 2 at 11:23
Asaf Karagila♦Asaf Karagila
302k32426757
302k32426757
add a comment |
add a comment |
Ali Bayati is a new contributor. Be nice, and check out our Code of Conduct.
Ali Bayati is a new contributor. Be nice, and check out our Code of Conduct.
Ali Bayati is a new contributor. Be nice, and check out our Code of Conduct.
Ali Bayati is a new contributor. Be nice, and check out our Code of Conduct.
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4
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.
– Georges Elencwajg
Jan 2 at 11:09
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.)
– Not Mike
Jan 2 at 11:20
I just want to know if this problem is open or it solve previously?
– Ali Bayati
Jan 2 at 11:23
If I had to guess, I'd guess this was solved close to a 100 years ago.
– Asaf Karagila♦
Jan 2 at 11:24
Are there any references in this area? Why we have $card( l^2(sigma))=card (sigma)$ when $card( sigma) geq c?$
– Ali Bayati
Jan 2 at 11:27