Mixing
Confusion about real separable normed space problem
$begingroup$
Suppose that $X$ is a real separable vector space, and $W$ a closed linear
subspace of $X$. Show that there exists a sequence $(z_j )in X$ such that
$z_{j+1} notin W_j := mathrm{span} W cup {z_1, . . . , z_j}$
and if we define $W_infty = mathrm{span} W cup {z_j}_{j=1}^infty$
then $overline{W_infty} = X$.
This is a problem in a (non-credit) example sheet I am working on currently. After spending some time trying it I have realised that it is probably incorrect as it stands, but I am not sure. I think that (even with the necessary assumption that $W$ is a proper subspace), $W=mathbb{C}$ (over $mathbb{R}$) is a counterexample? Maybe there is some implicit assumption in the problem that I am missing. I would appreciate if someone could clarify what the correct version of the problem would be. Please do not give me answers to the problem though, as I would like to first try it myself.
functional-analysis normed-spaces separable-spaces
asked Feb 2 at 14:51
AlephNullAlephNull
559110
$endgroup$
add a comment |
$begingroup$
Suppose that $X$ is a real separable vector space, and $W$ a closed linear
subspace of $X$. Show that there exists a sequence $(z_j )in X$ such that
$z_{j+1} notin W_j := mathrm{span} W cup {z_1, . . . , z_j}$
and if we define $W_infty = mathrm{span} W cup {z_j}_{j=1}^infty$
then $overline{W_infty} = X$.
This is a problem in a (non-credit) example sheet I am working on currently. After spending some time trying it I have realised that it is probably incorrect as it stands, but I am not sure. I think that (even with the necessary assumption that $W$ is a proper subspace), $W=mathbb{C}$ (over $mathbb{R}$) is a counterexample? Maybe there is some implicit assumption in the problem that I am missing. I would appreciate if someone could clarify what the correct version of the problem would be. Please do not give me answers to the problem though, as I would like to first try it myself.
functional-analysis normed-spaces separable-spaces
asked Feb 2 at 14:51
AlephNullAlephNull
559110
$endgroup$
$begingroup$
Try adding the assumption that $X$ is infinite-diemensional.
$endgroup$
– David C. Ullrich
Feb 2 at 14:56
add a comment |
$begingroup$
Suppose that $X$ is a real separable vector space, and $W$ a closed linear
subspace of $X$. Show that there exists a sequence $(z_j )in X$ such that
$z_{j+1} notin W_j := mathrm{span} W cup {z_1, . . . , z_j}$
and if we define $W_infty = mathrm{span} W cup {z_j}_{j=1}^infty$
then $overline{W_infty} = X$.
This is a problem in a (non-credit) example sheet I am working on currently. After spending some time trying it I have realised that it is probably incorrect as it stands, but I am not sure. I think that (even with the necessary assumption that $W$ is a proper subspace), $W=mathbb{C}$ (over $mathbb{R}$) is a counterexample? Maybe there is some implicit assumption in the problem that I am missing. I would appreciate if someone could clarify what the correct version of the problem would be. Please do not give me answers to the problem though, as I would like to first try it myself.
functional-analysis normed-spaces separable-spaces
asked Feb 2 at 14:51
AlephNullAlephNull
559110
$endgroup$
Suppose that $X$ is a real separable vector space, and $W$ a closed linear
subspace of $X$. Show that there exists a sequence $(z_j )in X$ such that
$z_{j+1} notin W_j := mathrm{span} W cup {z_1, . . . , z_j}$
and if we define $W_infty = mathrm{span} W cup {z_j}_{j=1}^infty$
then $overline{W_infty} = X$.
This is a problem in a (non-credit) example sheet I am working on currently. After spending some time trying it I have realised that it is probably incorrect as it stands, but I am not sure. I think that (even with the necessary assumption that $W$ is a proper subspace), $W=mathbb{C}$ (over $mathbb{R}$) is a counterexample? Maybe there is some implicit assumption in the problem that I am missing. I would appreciate if someone could clarify what the correct version of the problem would be. Please do not give me answers to the problem though, as I would like to first try it myself.
functional-analysis normed-spaces separable-spaces
functional-analysis normed-spaces separable-spaces
asked Feb 2 at 14:51
AlephNullAlephNull
559110
asked Feb 2 at 14:51
AlephNullAlephNull
559110
asked Feb 2 at 14:51
AlephNullAlephNull
559110
asked Feb 2 at 14:51
AlephNullAlephNull
559110
asked Feb 2 at 14:51
AlephNullAlephNull
559110
559110
$begingroup$
Try adding the assumption that $X$ is infinite-diemensional.
$endgroup$
– David C. Ullrich
Feb 2 at 14:56
add a comment |
$begingroup$
Try adding the assumption that $X$ is infinite-diemensional.
$endgroup$
– David C. Ullrich
Feb 2 at 14:56
$begingroup$
Try adding the assumption that $X$ is infinite-diemensional.
$endgroup$
– David C. Ullrich
Feb 2 at 14:56
$begingroup$
Try adding the assumption that $X$ is infinite-diemensional.
$endgroup$
– David C. Ullrich
Feb 2 at 14:56
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
As stated, the result is false even for infinite-dimensional $X$. Because if for instance the codimension of $W$ is finite (say the kernel of a bounded linear functional, which has codimension 1), then no such sequence exists.
The easiest way to make the problem solvable would be to take $W$ finite-dimensional (of course you need $X$ infinite-dimensional, as the sequence ${z_j}$ is linearly independent). But that's clearly not the intention, since then it would irrelevant to say that $W$ is closed (as all finite-dimensional subspaces are). The other option I see is to say out right that the codimension of $W$ is infinite.
answered Feb 2 at 15:19
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
$begingroup$
I agree with all the points you made. That the codimension of $W$ is infinite is quite a strange assumption to be implicit, since the following question asks us to prove the Hahn-Banach Theorem for a real separable normed space using this result (together with the results that one can make such extensions for each addition of a $z_j$ and that we can extend to the closure). Edit: Actually it is not clear how it would follow with there being infinitely many $z_j$. I suppose I will just have to wait for the solutions to come out/the lecturer to clarify the problems.
$endgroup$
– AlephNull
Feb 2 at 15:50
1
$begingroup$
Yes, and actually there are even more hidden assumptions. If $W$ is separable and $X$ isn't, then you cannot get a sequence as in your question (because $overline{W_infty}$ is separable). That's precisely the reason one has to use Zorn's Lemma for Hahn-Banach, and not just something simpler like induction.
$endgroup$
– Martin Argerami
Feb 2 at 16:16
add a comment |
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1 Answer
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$begingroup$
As stated, the result is false even for infinite-dimensional $X$. Because if for instance the codimension of $W$ is finite (say the kernel of a bounded linear functional, which has codimension 1), then no such sequence exists.
The easiest way to make the problem solvable would be to take $W$ finite-dimensional (of course you need $X$ infinite-dimensional, as the sequence ${z_j}$ is linearly independent). But that's clearly not the intention, since then it would irrelevant to say that $W$ is closed (as all finite-dimensional subspaces are). The other option I see is to say out right that the codimension of $W$ is infinite.
answered Feb 2 at 15:19
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
$begingroup$
I agree with all the points you made. That the codimension of $W$ is infinite is quite a strange assumption to be implicit, since the following question asks us to prove the Hahn-Banach Theorem for a real separable normed space using this result (together with the results that one can make such extensions for each addition of a $z_j$ and that we can extend to the closure). Edit: Actually it is not clear how it would follow with there being infinitely many $z_j$. I suppose I will just have to wait for the solutions to come out/the lecturer to clarify the problems.
$endgroup$
– AlephNull
Feb 2 at 15:50
1
$begingroup$
Yes, and actually there are even more hidden assumptions. If $W$ is separable and $X$ isn't, then you cannot get a sequence as in your question (because $overline{W_infty}$ is separable). That's precisely the reason one has to use Zorn's Lemma for Hahn-Banach, and not just something simpler like induction.
$endgroup$
– Martin Argerami
Feb 2 at 16:16
add a comment |
$begingroup$
As stated, the result is false even for infinite-dimensional $X$. Because if for instance the codimension of $W$ is finite (say the kernel of a bounded linear functional, which has codimension 1), then no such sequence exists.
The easiest way to make the problem solvable would be to take $W$ finite-dimensional (of course you need $X$ infinite-dimensional, as the sequence ${z_j}$ is linearly independent). But that's clearly not the intention, since then it would irrelevant to say that $W$ is closed (as all finite-dimensional subspaces are). The other option I see is to say out right that the codimension of $W$ is infinite.
answered Feb 2 at 15:19
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
$begingroup$
I agree with all the points you made. That the codimension of $W$ is infinite is quite a strange assumption to be implicit, since the following question asks us to prove the Hahn-Banach Theorem for a real separable normed space using this result (together with the results that one can make such extensions for each addition of a $z_j$ and that we can extend to the closure). Edit: Actually it is not clear how it would follow with there being infinitely many $z_j$. I suppose I will just have to wait for the solutions to come out/the lecturer to clarify the problems.
$endgroup$
– AlephNull
Feb 2 at 15:50
1
$begingroup$
Yes, and actually there are even more hidden assumptions. If $W$ is separable and $X$ isn't, then you cannot get a sequence as in your question (because $overline{W_infty}$ is separable). That's precisely the reason one has to use Zorn's Lemma for Hahn-Banach, and not just something simpler like induction.
$endgroup$
– Martin Argerami
Feb 2 at 16:16
add a comment |
$begingroup$
As stated, the result is false even for infinite-dimensional $X$. Because if for instance the codimension of $W$ is finite (say the kernel of a bounded linear functional, which has codimension 1), then no such sequence exists.
The easiest way to make the problem solvable would be to take $W$ finite-dimensional (of course you need $X$ infinite-dimensional, as the sequence ${z_j}$ is linearly independent). But that's clearly not the intention, since then it would irrelevant to say that $W$ is closed (as all finite-dimensional subspaces are). The other option I see is to say out right that the codimension of $W$ is infinite.
answered Feb 2 at 15:19
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
As stated, the result is false even for infinite-dimensional $X$. Because if for instance the codimension of $W$ is finite (say the kernel of a bounded linear functional, which has codimension 1), then no such sequence exists.
The easiest way to make the problem solvable would be to take $W$ finite-dimensional (of course you need $X$ infinite-dimensional, as the sequence ${z_j}$ is linearly independent). But that's clearly not the intention, since then it would irrelevant to say that $W$ is closed (as all finite-dimensional subspaces are). The other option I see is to say out right that the codimension of $W$ is infinite.
answered Feb 2 at 15:19
Martin ArgeramiMartin Argerami
129k1184185
answered Feb 2 at 15:19
Martin ArgeramiMartin Argerami
129k1184185
answered Feb 2 at 15:19
Martin ArgeramiMartin Argerami
129k1184185
answered Feb 2 at 15:19
Martin ArgeramiMartin Argerami
129k1184185
129k1184185
$begingroup$
I agree with all the points you made. That the codimension of $W$ is infinite is quite a strange assumption to be implicit, since the following question asks us to prove the Hahn-Banach Theorem for a real separable normed space using this result (together with the results that one can make such extensions for each addition of a $z_j$ and that we can extend to the closure). Edit: Actually it is not clear how it would follow with there being infinitely many $z_j$. I suppose I will just have to wait for the solutions to come out/the lecturer to clarify the problems.
$endgroup$
– AlephNull
Feb 2 at 15:50
1
$begingroup$
Yes, and actually there are even more hidden assumptions. If $W$ is separable and $X$ isn't, then you cannot get a sequence as in your question (because $overline{W_infty}$ is separable). That's precisely the reason one has to use Zorn's Lemma for Hahn-Banach, and not just something simpler like induction.
$endgroup$
– Martin Argerami
Feb 2 at 16:16
add a comment |
$begingroup$
I agree with all the points you made. That the codimension of $W$ is infinite is quite a strange assumption to be implicit, since the following question asks us to prove the Hahn-Banach Theorem for a real separable normed space using this result (together with the results that one can make such extensions for each addition of a $z_j$ and that we can extend to the closure). Edit: Actually it is not clear how it would follow with there being infinitely many $z_j$. I suppose I will just have to wait for the solutions to come out/the lecturer to clarify the problems.
$endgroup$
– AlephNull
Feb 2 at 15:50
1
$begingroup$
Yes, and actually there are even more hidden assumptions. If $W$ is separable and $X$ isn't, then you cannot get a sequence as in your question (because $overline{W_infty}$ is separable). That's precisely the reason one has to use Zorn's Lemma for Hahn-Banach, and not just something simpler like induction.
$endgroup$
– Martin Argerami
Feb 2 at 16:16
$begingroup$
I agree with all the points you made. That the codimension of $W$ is infinite is quite a strange assumption to be implicit, since the following question asks us to prove the Hahn-Banach Theorem for a real separable normed space using this result (together with the results that one can make such extensions for each addition of a $z_j$ and that we can extend to the closure). Edit: Actually it is not clear how it would follow with there being infinitely many $z_j$. I suppose I will just have to wait for the solutions to come out/the lecturer to clarify the problems.
$endgroup$
– AlephNull
Feb 2 at 15:50
$begingroup$
I agree with all the points you made. That the codimension of $W$ is infinite is quite a strange assumption to be implicit, since the following question asks us to prove the Hahn-Banach Theorem for a real separable normed space using this result (together with the results that one can make such extensions for each addition of a $z_j$ and that we can extend to the closure). Edit: Actually it is not clear how it would follow with there being infinitely many $z_j$. I suppose I will just have to wait for the solutions to come out/the lecturer to clarify the problems.
$endgroup$
– AlephNull
Feb 2 at 15:50
1
1
$begingroup$
Yes, and actually there are even more hidden assumptions. If $W$ is separable and $X$ isn't, then you cannot get a sequence as in your question (because $overline{W_infty}$ is separable). That's precisely the reason one has to use Zorn's Lemma for Hahn-Banach, and not just something simpler like induction.
$endgroup$
– Martin Argerami
Feb 2 at 16:16
$begingroup$
Yes, and actually there are even more hidden assumptions. If $W$ is separable and $X$ isn't, then you cannot get a sequence as in your question (because $overline{W_infty}$ is separable). That's precisely the reason one has to use Zorn's Lemma for Hahn-Banach, and not just something simpler like induction.
$endgroup$
– Martin Argerami
Feb 2 at 16:16
add a comment |
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$begingroup$
Try adding the assumption that $X$ is infinite-diemensional.
$endgroup$
– David C. Ullrich
Feb 2 at 14:56