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Basis and cardinality
Uncountable basis and separabilityWhat is the difference between a Hamel basis and a Schauder basis?little question; nonseperable Hilbert spaces: what kind of basis…?Why isn't every Hamel basis a Schauder basis?Confusion about the Hamel BasisDoes every (non-separable) Hilbert space have the approximation property?An orthogonal basis of a Hilbert space is Schauder?Schauder basis that is not Hilbert basis“Basis” of non-separable Hilbert spaceUncountable basis in Hilbert space vs orthonormal basis
$begingroup$
Suppose that $B$ is a separable Hilbert space and $b_i_i in mathbbN$ is a countable linearly independent set. Then is $b_i_i in mathbbN$ necessarily a Schauder basis for $B$?
linear-algebra hilbert-spaces topological-vector-spaces schauder-basis hamel-basis
asked 9 hours ago
N00berN00ber
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add a comment |
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Suppose that $B$ is a separable Hilbert space and $b_i_i in mathbbN$ is a countable linearly independent set. Then is $b_i_i in mathbbN$ necessarily a Schauder basis for $B$?
linear-algebra hilbert-spaces topological-vector-spaces schauder-basis hamel-basis
asked 9 hours ago
N00berN00ber
4062 silver badges11 bronze badges
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add a comment |
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Suppose that $B$ is a separable Hilbert space and $b_i_i in mathbbN$ is a countable linearly independent set. Then is $b_i_i in mathbbN$ necessarily a Schauder basis for $B$?
linear-algebra hilbert-spaces topological-vector-spaces schauder-basis hamel-basis
asked 9 hours ago
N00berN00ber
4062 silver badges11 bronze badges
$endgroup$
Suppose that $B$ is a separable Hilbert space and $b_i_i in mathbbN$ is a countable linearly independent set. Then is $b_i_i in mathbbN$ necessarily a Schauder basis for $B$?
linear-algebra hilbert-spaces topological-vector-spaces schauder-basis hamel-basis
linear-algebra hilbert-spaces topological-vector-spaces schauder-basis hamel-basis
asked 9 hours ago
N00berN00ber
4062 silver badges11 bronze badges
asked 9 hours ago
N00berN00ber
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edited 7 hours ago
N00ber
asked 9 hours ago
N00berN00ber
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asked 9 hours ago
N00berN00ber
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asked 9 hours ago
N00berN00ber
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3 Answers
3
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Obviously not, because for example any subset is also independent, but a proper subset of a basis cannot be a basis.
answered 8 hours ago
David C. UllrichDavid C. Ullrich
63.8k4 gold badges44 silver badges99 bronze badges
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Sorry there wsa a typo. I mean't a countable linearly independent set.
$endgroup$
– N00ber
7 hours ago
$begingroup$
@N00ber ??? It was clear that you were talking about a countable independent set - what does that have to do with the answer I gave?
$endgroup$
– David C. Ullrich
7 hours ago
$begingroup$
Oh I tried to accept both but that can't be done. It's good now :)
$endgroup$
– N00ber
7 hours ago
add a comment |
$begingroup$
No. For example, the sequence
$$beginalign b_0 &= (1,0,0,0,0,dots)\ b_1 &= (0,0,1,0,0,dots)\ b_2 &= (0,0,0,0,1,dots)\ vdots &= vdots endalign$$
is linearly independent in $ell^2$, but not a Schauder basis. ($b_n$ has a $1$ in position $2n$ and zeros elsewhere.)
answered 8 hours ago
Alex KruckmanAlex Kruckman
30.8k3 gold badges29 silver badges60 bronze badges
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Very clear answer :)
$endgroup$
– N00ber
7 hours ago
add a comment |
$begingroup$
Besides taking a proper subset of a Schauder basis, you can also take a superset that is still (algebraically) linearly independent. For instance, if $(e_n)_ninmathbbN$ is an orthonormal basis, then $e_n_ninmathbbNcupv$ is still linearly independent where $v=sum_n e_n/2^n$, but is not a Schauder basis since the representation of $v$ as an infinite linear combination of basis elements is not unique.
answered 8 hours ago
Eric WofseyEric Wofsey
201k14 gold badges234 silver badges365 bronze badges
$endgroup$
$begingroup$
Wait, so every countable linearly-independent subset of $B$ cannot be "extended" to a Schauder basis of $B$?
$endgroup$
– N00ber
5 hours ago
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Obviously not, because for example any subset is also independent, but a proper subset of a basis cannot be a basis.
answered 8 hours ago
David C. UllrichDavid C. Ullrich
63.8k4 gold badges44 silver badges99 bronze badges
$endgroup$
$begingroup$
Sorry there wsa a typo. I mean't a countable linearly independent set.
$endgroup$
– N00ber
7 hours ago
$begingroup$
@N00ber ??? It was clear that you were talking about a countable independent set - what does that have to do with the answer I gave?
$endgroup$
– David C. Ullrich
7 hours ago
$begingroup$
Oh I tried to accept both but that can't be done. It's good now :)
$endgroup$
– N00ber
7 hours ago
add a comment |
$begingroup$
Obviously not, because for example any subset is also independent, but a proper subset of a basis cannot be a basis.
answered 8 hours ago
David C. UllrichDavid C. Ullrich
63.8k4 gold badges44 silver badges99 bronze badges
$endgroup$
$begingroup$
Sorry there wsa a typo. I mean't a countable linearly independent set.
$endgroup$
– N00ber
7 hours ago
$begingroup$
@N00ber ??? It was clear that you were talking about a countable independent set - what does that have to do with the answer I gave?
$endgroup$
– David C. Ullrich
7 hours ago
$begingroup$
Oh I tried to accept both but that can't be done. It's good now :)
$endgroup$
– N00ber
7 hours ago
add a comment |
$begingroup$
Obviously not, because for example any subset is also independent, but a proper subset of a basis cannot be a basis.
answered 8 hours ago
David C. UllrichDavid C. Ullrich
63.8k4 gold badges44 silver badges99 bronze badges
$endgroup$
Obviously not, because for example any subset is also independent, but a proper subset of a basis cannot be a basis.
answered 8 hours ago
David C. UllrichDavid C. Ullrich
63.8k4 gold badges44 silver badges99 bronze badges
answered 8 hours ago
David C. UllrichDavid C. Ullrich
63.8k4 gold badges44 silver badges99 bronze badges
answered 8 hours ago
David C. UllrichDavid C. Ullrich
63.8k4 gold badges44 silver badges99 bronze badges
answered 8 hours ago
David C. UllrichDavid C. Ullrich
63.8k4 gold badges44 silver badges99 bronze badges
63.8k4 gold badges44 silver badges99 bronze badges
$begingroup$
Sorry there wsa a typo. I mean't a countable linearly independent set.
$endgroup$
– N00ber
7 hours ago
$begingroup$
@N00ber ??? It was clear that you were talking about a countable independent set - what does that have to do with the answer I gave?
$endgroup$
– David C. Ullrich
7 hours ago
$begingroup$
Oh I tried to accept both but that can't be done. It's good now :)
$endgroup$
– N00ber
7 hours ago
add a comment |
$begingroup$
Sorry there wsa a typo. I mean't a countable linearly independent set.
$endgroup$
– N00ber
7 hours ago
$begingroup$
@N00ber ??? It was clear that you were talking about a countable independent set - what does that have to do with the answer I gave?
$endgroup$
– David C. Ullrich
7 hours ago
$begingroup$
Oh I tried to accept both but that can't be done. It's good now :)
$endgroup$
– N00ber
7 hours ago
$begingroup$
Sorry there wsa a typo. I mean't a countable linearly independent set.
$endgroup$
– N00ber
7 hours ago
$begingroup$
Sorry there wsa a typo. I mean't a countable linearly independent set.
$endgroup$
– N00ber
7 hours ago
$begingroup$
@N00ber ??? It was clear that you were talking about a countable independent set - what does that have to do with the answer I gave?
$endgroup$
– David C. Ullrich
7 hours ago
$begingroup$
@N00ber ??? It was clear that you were talking about a countable independent set - what does that have to do with the answer I gave?
$endgroup$
– David C. Ullrich
7 hours ago
$begingroup$
Oh I tried to accept both but that can't be done. It's good now :)
$endgroup$
– N00ber
7 hours ago
$begingroup$
Oh I tried to accept both but that can't be done. It's good now :)
$endgroup$
– N00ber
7 hours ago
add a comment |
$begingroup$
No. For example, the sequence
$$beginalign b_0 &= (1,0,0,0,0,dots)\ b_1 &= (0,0,1,0,0,dots)\ b_2 &= (0,0,0,0,1,dots)\ vdots &= vdots endalign$$
is linearly independent in $ell^2$, but not a Schauder basis. ($b_n$ has a $1$ in position $2n$ and zeros elsewhere.)
answered 8 hours ago
Alex KruckmanAlex Kruckman
30.8k3 gold badges29 silver badges60 bronze badges
$endgroup$
$begingroup$
Very clear answer :)
$endgroup$
– N00ber
7 hours ago
add a comment |
$begingroup$
No. For example, the sequence
$$beginalign b_0 &= (1,0,0,0,0,dots)\ b_1 &= (0,0,1,0,0,dots)\ b_2 &= (0,0,0,0,1,dots)\ vdots &= vdots endalign$$
is linearly independent in $ell^2$, but not a Schauder basis. ($b_n$ has a $1$ in position $2n$ and zeros elsewhere.)
answered 8 hours ago
Alex KruckmanAlex Kruckman
30.8k3 gold badges29 silver badges60 bronze badges
$endgroup$
$begingroup$
Very clear answer :)
$endgroup$
– N00ber
7 hours ago
add a comment |
$begingroup$
No. For example, the sequence
$$beginalign b_0 &= (1,0,0,0,0,dots)\ b_1 &= (0,0,1,0,0,dots)\ b_2 &= (0,0,0,0,1,dots)\ vdots &= vdots endalign$$
is linearly independent in $ell^2$, but not a Schauder basis. ($b_n$ has a $1$ in position $2n$ and zeros elsewhere.)
answered 8 hours ago
Alex KruckmanAlex Kruckman
30.8k3 gold badges29 silver badges60 bronze badges
$endgroup$
No. For example, the sequence
$$beginalign b_0 &= (1,0,0,0,0,dots)\ b_1 &= (0,0,1,0,0,dots)\ b_2 &= (0,0,0,0,1,dots)\ vdots &= vdots endalign$$
is linearly independent in $ell^2$, but not a Schauder basis. ($b_n$ has a $1$ in position $2n$ and zeros elsewhere.)
answered 8 hours ago
Alex KruckmanAlex Kruckman
30.8k3 gold badges29 silver badges60 bronze badges
answered 8 hours ago
Alex KruckmanAlex Kruckman
30.8k3 gold badges29 silver badges60 bronze badges
answered 8 hours ago
Alex KruckmanAlex Kruckman
30.8k3 gold badges29 silver badges60 bronze badges
answered 8 hours ago
Alex KruckmanAlex Kruckman
30.8k3 gold badges29 silver badges60 bronze badges
30.8k3 gold badges29 silver badges60 bronze badges
$begingroup$
Very clear answer :)
$endgroup$
– N00ber
7 hours ago
add a comment |
$begingroup$
Very clear answer :)
$endgroup$
– N00ber
7 hours ago
$begingroup$
Very clear answer :)
$endgroup$
– N00ber
7 hours ago
$begingroup$
Very clear answer :)
$endgroup$
– N00ber
7 hours ago
add a comment |
$begingroup$
Besides taking a proper subset of a Schauder basis, you can also take a superset that is still (algebraically) linearly independent. For instance, if $(e_n)_ninmathbbN$ is an orthonormal basis, then $e_n_ninmathbbNcupv$ is still linearly independent where $v=sum_n e_n/2^n$, but is not a Schauder basis since the representation of $v$ as an infinite linear combination of basis elements is not unique.
answered 8 hours ago
Eric WofseyEric Wofsey
201k14 gold badges234 silver badges365 bronze badges
$endgroup$
$begingroup$
Wait, so every countable linearly-independent subset of $B$ cannot be "extended" to a Schauder basis of $B$?
$endgroup$
– N00ber
5 hours ago
add a comment |
$begingroup$
Besides taking a proper subset of a Schauder basis, you can also take a superset that is still (algebraically) linearly independent. For instance, if $(e_n)_ninmathbbN$ is an orthonormal basis, then $e_n_ninmathbbNcupv$ is still linearly independent where $v=sum_n e_n/2^n$, but is not a Schauder basis since the representation of $v$ as an infinite linear combination of basis elements is not unique.
answered 8 hours ago
Eric WofseyEric Wofsey
201k14 gold badges234 silver badges365 bronze badges
$endgroup$
$begingroup$
Wait, so every countable linearly-independent subset of $B$ cannot be "extended" to a Schauder basis of $B$?
$endgroup$
– N00ber
5 hours ago
add a comment |
$begingroup$
Besides taking a proper subset of a Schauder basis, you can also take a superset that is still (algebraically) linearly independent. For instance, if $(e_n)_ninmathbbN$ is an orthonormal basis, then $e_n_ninmathbbNcupv$ is still linearly independent where $v=sum_n e_n/2^n$, but is not a Schauder basis since the representation of $v$ as an infinite linear combination of basis elements is not unique.
answered 8 hours ago
Eric WofseyEric Wofsey
201k14 gold badges234 silver badges365 bronze badges
$endgroup$
Besides taking a proper subset of a Schauder basis, you can also take a superset that is still (algebraically) linearly independent. For instance, if $(e_n)_ninmathbbN$ is an orthonormal basis, then $e_n_ninmathbbNcupv$ is still linearly independent where $v=sum_n e_n/2^n$, but is not a Schauder basis since the representation of $v$ as an infinite linear combination of basis elements is not unique.
answered 8 hours ago
Eric WofseyEric Wofsey
201k14 gold badges234 silver badges365 bronze badges
answered 8 hours ago
Eric WofseyEric Wofsey
201k14 gold badges234 silver badges365 bronze badges
answered 8 hours ago
Eric WofseyEric Wofsey
201k14 gold badges234 silver badges365 bronze badges
answered 8 hours ago
Eric WofseyEric Wofsey
201k14 gold badges234 silver badges365 bronze badges
201k14 gold badges234 silver badges365 bronze badges
$begingroup$
Wait, so every countable linearly-independent subset of $B$ cannot be "extended" to a Schauder basis of $B$?
$endgroup$
– N00ber
5 hours ago
add a comment |
$begingroup$
Wait, so every countable linearly-independent subset of $B$ cannot be "extended" to a Schauder basis of $B$?
$endgroup$
– N00ber
5 hours ago
$begingroup$
Wait, so every countable linearly-independent subset of $B$ cannot be "extended" to a Schauder basis of $B$?
$endgroup$
– N00ber
5 hours ago
$begingroup$
Wait, so every countable linearly-independent subset of $B$ cannot be "extended" to a Schauder basis of $B$?
$endgroup$
– N00ber
5 hours ago
add a comment |
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