Mixing
Finding a basis for the vector space $mathbb{R}$ over $mathbb{Q}$ [closed]
$begingroup$
Consider the vector space $mathbb{R}$ over $mathbb{Q}$. How can I prove that ${1,pi,pi^2, ldots}$ is a basis for this space?
vector-spaces
edited Jan 12 at 1:58
amWhy
1
asked Jan 12 at 1:37
HiggsinoHiggsino
595
$endgroup$
closed as unclear what you're asking by Saad, metamorphy, user91500, Arnaud D., Paul Frost Jan 12 at 23:04
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Consider the vector space $mathbb{R}$ over $mathbb{Q}$. How can I prove that ${1,pi,pi^2, ldots}$ is a basis for this space?
vector-spaces
edited Jan 12 at 1:58
amWhy
1
asked Jan 12 at 1:37
HiggsinoHiggsino
595
$endgroup$
closed as unclear what you're asking by Saad, metamorphy, user91500, Arnaud D., Paul Frost Jan 12 at 23:04
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
4
$begingroup$
The set ${1,pi,pi^2,ldots}$ is not a basis for $mathbf{R}$ over $mathbf{Q}$. It is a $mathbf{Q}$-linearly independent set that cannot span $mathbf{R}$ as a $mathbf{Q}$-vector space for cardinality reasons.
$endgroup$
– Keenan Kidwell
Jan 12 at 1:39
4
$begingroup$
Who said this was a basis?
$endgroup$
– Randall
Jan 12 at 1:40
$begingroup$
Yeah, probably my mistake then. BUt if this is not a basis, then does another basis exist?
$endgroup$
– Higgsino
Jan 12 at 1:42
$begingroup$
mathworld.wolfram.com/HamelBasis.html
$endgroup$
– Bungo
Jan 12 at 1:59
$begingroup$
Considering the answer is something along the lines of “you can’t” / “it’s not,” I’m wondering where this question came from?
$endgroup$
– Chase Ryan Taylor
Jan 12 at 5:26
add a comment |
$begingroup$
Consider the vector space $mathbb{R}$ over $mathbb{Q}$. How can I prove that ${1,pi,pi^2, ldots}$ is a basis for this space?
vector-spaces
edited Jan 12 at 1:58
amWhy
1
asked Jan 12 at 1:37
HiggsinoHiggsino
595
$endgroup$
Consider the vector space $mathbb{R}$ over $mathbb{Q}$. How can I prove that ${1,pi,pi^2, ldots}$ is a basis for this space?
vector-spaces
vector-spaces
edited Jan 12 at 1:58
amWhy
1
asked Jan 12 at 1:37
HiggsinoHiggsino
595
edited Jan 12 at 1:58
amWhy
1
asked Jan 12 at 1:37
HiggsinoHiggsino
595
edited Jan 12 at 1:58
amWhy
1
edited Jan 12 at 1:58
amWhy
1
edited Jan 12 at 1:58
amWhy
1
1
asked Jan 12 at 1:37
HiggsinoHiggsino
595
asked Jan 12 at 1:37
HiggsinoHiggsino
595
asked Jan 12 at 1:37
HiggsinoHiggsino
595
595
closed as unclear what you're asking by Saad, metamorphy, user91500, Arnaud D., Paul Frost Jan 12 at 23:04
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Saad, metamorphy, user91500, Arnaud D., Paul Frost Jan 12 at 23:04
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
4
$begingroup$
The set ${1,pi,pi^2,ldots}$ is not a basis for $mathbf{R}$ over $mathbf{Q}$. It is a $mathbf{Q}$-linearly independent set that cannot span $mathbf{R}$ as a $mathbf{Q}$-vector space for cardinality reasons.
$endgroup$
– Keenan Kidwell
Jan 12 at 1:39
4
$begingroup$
Who said this was a basis?
$endgroup$
– Randall
Jan 12 at 1:40
$begingroup$
Yeah, probably my mistake then. BUt if this is not a basis, then does another basis exist?
$endgroup$
– Higgsino
Jan 12 at 1:42
$begingroup$
mathworld.wolfram.com/HamelBasis.html
$endgroup$
– Bungo
Jan 12 at 1:59
$begingroup$
Considering the answer is something along the lines of “you can’t” / “it’s not,” I’m wondering where this question came from?
$endgroup$
– Chase Ryan Taylor
Jan 12 at 5:26
add a comment |
4
$begingroup$
The set ${1,pi,pi^2,ldots}$ is not a basis for $mathbf{R}$ over $mathbf{Q}$. It is a $mathbf{Q}$-linearly independent set that cannot span $mathbf{R}$ as a $mathbf{Q}$-vector space for cardinality reasons.
$endgroup$
– Keenan Kidwell
Jan 12 at 1:39
4
$begingroup$
Who said this was a basis?
$endgroup$
– Randall
Jan 12 at 1:40
$begingroup$
Yeah, probably my mistake then. BUt if this is not a basis, then does another basis exist?
$endgroup$
– Higgsino
Jan 12 at 1:42
$begingroup$
mathworld.wolfram.com/HamelBasis.html
$endgroup$
– Bungo
Jan 12 at 1:59
$begingroup$
Considering the answer is something along the lines of “you can’t” / “it’s not,” I’m wondering where this question came from?
$endgroup$
– Chase Ryan Taylor
Jan 12 at 5:26
4
4
$begingroup$
The set ${1,pi,pi^2,ldots}$ is not a basis for $mathbf{R}$ over $mathbf{Q}$. It is a $mathbf{Q}$-linearly independent set that cannot span $mathbf{R}$ as a $mathbf{Q}$-vector space for cardinality reasons.
$endgroup$
– Keenan Kidwell
Jan 12 at 1:39
$begingroup$
The set ${1,pi,pi^2,ldots}$ is not a basis for $mathbf{R}$ over $mathbf{Q}$. It is a $mathbf{Q}$-linearly independent set that cannot span $mathbf{R}$ as a $mathbf{Q}$-vector space for cardinality reasons.
$endgroup$
– Keenan Kidwell
Jan 12 at 1:39
4
4
$begingroup$
Who said this was a basis?
$endgroup$
– Randall
Jan 12 at 1:40
$begingroup$
Who said this was a basis?
$endgroup$
– Randall
Jan 12 at 1:40
$begingroup$
Yeah, probably my mistake then. BUt if this is not a basis, then does another basis exist?
$endgroup$
– Higgsino
Jan 12 at 1:42
$begingroup$
Yeah, probably my mistake then. BUt if this is not a basis, then does another basis exist?
$endgroup$
– Higgsino
Jan 12 at 1:42
$begingroup$
mathworld.wolfram.com/HamelBasis.html
$endgroup$
– Bungo
Jan 12 at 1:59
$begingroup$
mathworld.wolfram.com/HamelBasis.html
$endgroup$
– Bungo
Jan 12 at 1:59
$begingroup$
Considering the answer is something along the lines of “you can’t” / “it’s not,” I’m wondering where this question came from?
$endgroup$
– Chase Ryan Taylor
Jan 12 at 5:26
$begingroup$
Considering the answer is something along the lines of “you can’t” / “it’s not,” I’m wondering where this question came from?
$endgroup$
– Chase Ryan Taylor
Jan 12 at 5:26
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
One may prove, using the Axiom of Choice, that a basis for $mathbb R$ over $mathbb Q$ exists. However that in itself will not allow you to write down the elements of such a basis.
The use of the Axiom of Choice is generally thought to be essential: Under quite mild set-theoretic assumptions, there are models of standard set theory without the Axiom of Choice where $mathbb R$ does not have a basis over $mathbb Q$. This is the case, for example, in the Solovay model. In that model all subsets of $mathbb R$ are Lebesgue measurable, but a $mathbb Q$-basis for $mathbb R$ would imply the existence of a Vitali set, which cannot be measurable.
This means that a basis cannot even be described in terms that make sense in the absense of AC -- but that includes all notations that are usually considered to be "explicit".
answered Jan 12 at 1:48
Henning MakholmHenning Makholm
240k17305541
$endgroup$
$begingroup$
Solid answer. Learned something. +1
$endgroup$
– Randall
Jan 12 at 1:59
2
$begingroup$
(Now expecting Asaf to sweep in any moment and say that the "generally thought to" hedge is unjustified here since because of such-and-such you can always have models without Vitali sets but with certain other unmeasurable sets).
$endgroup$
– Henning Makholm
Jan 12 at 2:02
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
One may prove, using the Axiom of Choice, that a basis for $mathbb R$ over $mathbb Q$ exists. However that in itself will not allow you to write down the elements of such a basis.
The use of the Axiom of Choice is generally thought to be essential: Under quite mild set-theoretic assumptions, there are models of standard set theory without the Axiom of Choice where $mathbb R$ does not have a basis over $mathbb Q$. This is the case, for example, in the Solovay model. In that model all subsets of $mathbb R$ are Lebesgue measurable, but a $mathbb Q$-basis for $mathbb R$ would imply the existence of a Vitali set, which cannot be measurable.
This means that a basis cannot even be described in terms that make sense in the absense of AC -- but that includes all notations that are usually considered to be "explicit".
answered Jan 12 at 1:48
Henning MakholmHenning Makholm
240k17305541
$endgroup$
$begingroup$
Solid answer. Learned something. +1
$endgroup$
– Randall
Jan 12 at 1:59
2
$begingroup$
(Now expecting Asaf to sweep in any moment and say that the "generally thought to" hedge is unjustified here since because of such-and-such you can always have models without Vitali sets but with certain other unmeasurable sets).
$endgroup$
– Henning Makholm
Jan 12 at 2:02
add a comment |
$begingroup$
One may prove, using the Axiom of Choice, that a basis for $mathbb R$ over $mathbb Q$ exists. However that in itself will not allow you to write down the elements of such a basis.
The use of the Axiom of Choice is generally thought to be essential: Under quite mild set-theoretic assumptions, there are models of standard set theory without the Axiom of Choice where $mathbb R$ does not have a basis over $mathbb Q$. This is the case, for example, in the Solovay model. In that model all subsets of $mathbb R$ are Lebesgue measurable, but a $mathbb Q$-basis for $mathbb R$ would imply the existence of a Vitali set, which cannot be measurable.
This means that a basis cannot even be described in terms that make sense in the absense of AC -- but that includes all notations that are usually considered to be "explicit".
answered Jan 12 at 1:48
Henning MakholmHenning Makholm
240k17305541
$endgroup$
$begingroup$
Solid answer. Learned something. +1
$endgroup$
– Randall
Jan 12 at 1:59
2
$begingroup$
(Now expecting Asaf to sweep in any moment and say that the "generally thought to" hedge is unjustified here since because of such-and-such you can always have models without Vitali sets but with certain other unmeasurable sets).
$endgroup$
– Henning Makholm
Jan 12 at 2:02
add a comment |
$begingroup$
One may prove, using the Axiom of Choice, that a basis for $mathbb R$ over $mathbb Q$ exists. However that in itself will not allow you to write down the elements of such a basis.
The use of the Axiom of Choice is generally thought to be essential: Under quite mild set-theoretic assumptions, there are models of standard set theory without the Axiom of Choice where $mathbb R$ does not have a basis over $mathbb Q$. This is the case, for example, in the Solovay model. In that model all subsets of $mathbb R$ are Lebesgue measurable, but a $mathbb Q$-basis for $mathbb R$ would imply the existence of a Vitali set, which cannot be measurable.
This means that a basis cannot even be described in terms that make sense in the absense of AC -- but that includes all notations that are usually considered to be "explicit".
answered Jan 12 at 1:48
Henning MakholmHenning Makholm
240k17305541
$endgroup$
One may prove, using the Axiom of Choice, that a basis for $mathbb R$ over $mathbb Q$ exists. However that in itself will not allow you to write down the elements of such a basis.
The use of the Axiom of Choice is generally thought to be essential: Under quite mild set-theoretic assumptions, there are models of standard set theory without the Axiom of Choice where $mathbb R$ does not have a basis over $mathbb Q$. This is the case, for example, in the Solovay model. In that model all subsets of $mathbb R$ are Lebesgue measurable, but a $mathbb Q$-basis for $mathbb R$ would imply the existence of a Vitali set, which cannot be measurable.
This means that a basis cannot even be described in terms that make sense in the absense of AC -- but that includes all notations that are usually considered to be "explicit".
answered Jan 12 at 1:48
Henning MakholmHenning Makholm
240k17305541
edited Jan 12 at 1:54
answered Jan 12 at 1:48
Henning MakholmHenning Makholm
240k17305541
answered Jan 12 at 1:48
Henning MakholmHenning Makholm
240k17305541
answered Jan 12 at 1:48
Henning MakholmHenning Makholm
240k17305541
240k17305541
$begingroup$
Solid answer. Learned something. +1
$endgroup$
– Randall
Jan 12 at 1:59
2
$begingroup$
(Now expecting Asaf to sweep in any moment and say that the "generally thought to" hedge is unjustified here since because of such-and-such you can always have models without Vitali sets but with certain other unmeasurable sets).
$endgroup$
– Henning Makholm
Jan 12 at 2:02
add a comment |
$begingroup$
Solid answer. Learned something. +1
$endgroup$
– Randall
Jan 12 at 1:59
2
$begingroup$
(Now expecting Asaf to sweep in any moment and say that the "generally thought to" hedge is unjustified here since because of such-and-such you can always have models without Vitali sets but with certain other unmeasurable sets).
$endgroup$
– Henning Makholm
Jan 12 at 2:02
$begingroup$
Solid answer. Learned something. +1
$endgroup$
– Randall
Jan 12 at 1:59
$begingroup$
Solid answer. Learned something. +1
$endgroup$
– Randall
Jan 12 at 1:59
2
2
$begingroup$
(Now expecting Asaf to sweep in any moment and say that the "generally thought to" hedge is unjustified here since because of such-and-such you can always have models without Vitali sets but with certain other unmeasurable sets).
$endgroup$
– Henning Makholm
Jan 12 at 2:02
$begingroup$
(Now expecting Asaf to sweep in any moment and say that the "generally thought to" hedge is unjustified here since because of such-and-such you can always have models without Vitali sets but with certain other unmeasurable sets).
$endgroup$
– Henning Makholm
Jan 12 at 2:02
add a comment |
4
$begingroup$
The set ${1,pi,pi^2,ldots}$ is not a basis for $mathbf{R}$ over $mathbf{Q}$. It is a $mathbf{Q}$-linearly independent set that cannot span $mathbf{R}$ as a $mathbf{Q}$-vector space for cardinality reasons.
$endgroup$
– Keenan Kidwell
Jan 12 at 1:39
4
$begingroup$
Who said this was a basis?
$endgroup$
– Randall
Jan 12 at 1:40
$begingroup$
Yeah, probably my mistake then. BUt if this is not a basis, then does another basis exist?
$endgroup$
– Higgsino
Jan 12 at 1:42
$begingroup$
mathworld.wolfram.com/HamelBasis.html
$endgroup$
– Bungo
Jan 12 at 1:59
$begingroup$
Considering the answer is something along the lines of “you can’t” / “it’s not,” I’m wondering where this question came from?
$endgroup$
– Chase Ryan Taylor
Jan 12 at 5:26