Is it true that: $V$ is isomorphic to $U$ and $W$ $Longrightarrow$ $V$ isomorphic to $U + W$

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Let $V$,$U$ and $W$ be closed and infinite dimensional subspaces of $ell^2$ such that $V$ is isomorphic to $U$ and $W$



My question: Is $V$ isomorphic to $U + W$



Thanks.




functional-analysis hilbert-spaces banach-spaces




share|cite|improve this question





















  • What is the meaning of "$V$ is isomorphic to $U$ and $W$" ? Does it mean "$V$ is isomorphic to $U$ and $V$ is isomorphic to $W$"?
    – GEdgar
    Jul 22 at 19:02










  • @GEdgar yes it is
    – Matey Math
    Jul 22 at 19:03










  • By isomorphic you mean isometrically isomorphic, right?
    – mechanodroid
    Jul 22 at 19:33










  • @mechanodroid yes it is
    – Matey Math
    Jul 22 at 19:34














up vote
2
down vote

favorite












Let $V$,$U$ and $W$ be closed and infinite dimensional subspaces of $ell^2$ such that $V$ is isomorphic to $U$ and $W$



My question: Is $V$ isomorphic to $U + W$



Thanks.




functional-analysis hilbert-spaces banach-spaces




share|cite|improve this question





















  • What is the meaning of "$V$ is isomorphic to $U$ and $W$" ? Does it mean "$V$ is isomorphic to $U$ and $V$ is isomorphic to $W$"?
    – GEdgar
    Jul 22 at 19:02










  • @GEdgar yes it is
    – Matey Math
    Jul 22 at 19:03










  • By isomorphic you mean isometrically isomorphic, right?
    – mechanodroid
    Jul 22 at 19:33










  • @mechanodroid yes it is
    – Matey Math
    Jul 22 at 19:34












up vote
2
down vote

favorite









up vote
2
down vote

favorite











Let $V$,$U$ and $W$ be closed and infinite dimensional subspaces of $ell^2$ such that $V$ is isomorphic to $U$ and $W$



My question: Is $V$ isomorphic to $U + W$



Thanks.




functional-analysis hilbert-spaces banach-spaces




share|cite|improve this question













Let $V$,$U$ and $W$ be closed and infinite dimensional subspaces of $ell^2$ such that $V$ is isomorphic to $U$ and $W$



My question: Is $V$ isomorphic to $U + W$



Thanks.





functional-analysis hilbert-spaces banach-spaces





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 22 at 19:05
























asked Jul 22 at 18:53









Matey Math

827413




827413











  • What is the meaning of "$V$ is isomorphic to $U$ and $W$" ? Does it mean "$V$ is isomorphic to $U$ and $V$ is isomorphic to $W$"?
    – GEdgar
    Jul 22 at 19:02










  • @GEdgar yes it is
    – Matey Math
    Jul 22 at 19:03










  • By isomorphic you mean isometrically isomorphic, right?
    – mechanodroid
    Jul 22 at 19:33










  • @mechanodroid yes it is
    – Matey Math
    Jul 22 at 19:34
















  • What is the meaning of "$V$ is isomorphic to $U$ and $W$" ? Does it mean "$V$ is isomorphic to $U$ and $V$ is isomorphic to $W$"?
    – GEdgar
    Jul 22 at 19:02










  • @GEdgar yes it is
    – Matey Math
    Jul 22 at 19:03










  • By isomorphic you mean isometrically isomorphic, right?
    – mechanodroid
    Jul 22 at 19:33










  • @mechanodroid yes it is
    – Matey Math
    Jul 22 at 19:34















What is the meaning of "$V$ is isomorphic to $U$ and $W$" ? Does it mean "$V$ is isomorphic to $U$ and $V$ is isomorphic to $W$"?
– GEdgar
Jul 22 at 19:02




What is the meaning of "$V$ is isomorphic to $U$ and $W$" ? Does it mean "$V$ is isomorphic to $U$ and $V$ is isomorphic to $W$"?
– GEdgar
Jul 22 at 19:02












@GEdgar yes it is
– Matey Math
Jul 22 at 19:03




@GEdgar yes it is
– Matey Math
Jul 22 at 19:03












By isomorphic you mean isometrically isomorphic, right?
– mechanodroid
Jul 22 at 19:33




By isomorphic you mean isometrically isomorphic, right?
– mechanodroid
Jul 22 at 19:33












@mechanodroid yes it is
– Matey Math
Jul 22 at 19:34




@mechanodroid yes it is
– Matey Math
Jul 22 at 19:34










2 Answers
2






active

oldest

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up vote
1
down vote



accepted










The answer to the question as stated is no.



Let $V = overlineoperatornamespan e_2n_ninmathbbN$ and $U = overlineoperatornamespan lefte_2n + frace_2n+1n+1right_ninmathbbN$.



Then $U, V$ are infinite-dimensional closed subspaces of $ell^2$ and $V cong V$, $V cong U$. The latter are isometrically isomorphic via $$e_2n mapsto frac1sqrt1+frac1(n+1)^2 left(e_2n + frace_2n+1n+1right)$$



However, it is shown here that $V + U$ is not closed so it is not complete. But $V$ is complete so $V$ and $V + U$ cannot be isometrically isomorphic.






share|cite|improve this answer





















  • thanks for your answer
    – Matey Math
    Jul 22 at 19:46

















up vote
1
down vote













No in general, as sum of two closed subspace is not necessarily closed. See: The direct sum of two closed subspace is closed? (Hilbert space)



However, if $W + U$ happens to be closed, then both $V$ and $W + U$ will be isomorphic to $ell^2$ itself, as infinite dimensional subspace of separable Hilbert space is also a separable Hilbert space. Hence, $V cong ell^2 cong W + U$.






share|cite|improve this answer























  • now I edited that the spaces are infinite dimensional.
    – Matey Math
    Jul 22 at 19:06











  • In fact every Hilbert space with countable Schauder Basis (or eqv. separable) is isomorphic to $ell^2$. This is true for closed $infty$-dimensional subspace of $ell^2$. So the answer must be yes now.
    – Nik Pronko
    Jul 22 at 19:18











  • thanks @Nik Pronko for your answer P.S. if you put your comment in an answer i can sign it as answered
    – Matey Math
    Jul 22 at 19:20











  • No it still can be wrong as sum of two closed subspaces is not necessarily closed. See: math.stackexchange.com/questions/135471/…
    – Nik Pronko
    Jul 22 at 19:28






  • 1




    Yes, then all subspaces just isomorphic to $ell^2$.
    – Nik Pronko
    Jul 22 at 19:34










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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote



accepted










The answer to the question as stated is no.



Let $V = overlineoperatornamespan e_2n_ninmathbbN$ and $U = overlineoperatornamespan lefte_2n + frace_2n+1n+1right_ninmathbbN$.



Then $U, V$ are infinite-dimensional closed subspaces of $ell^2$ and $V cong V$, $V cong U$. The latter are isometrically isomorphic via $$e_2n mapsto frac1sqrt1+frac1(n+1)^2 left(e_2n + frace_2n+1n+1right)$$



However, it is shown here that $V + U$ is not closed so it is not complete. But $V$ is complete so $V$ and $V + U$ cannot be isometrically isomorphic.






share|cite|improve this answer





















  • thanks for your answer
    – Matey Math
    Jul 22 at 19:46














up vote
1
down vote



accepted










The answer to the question as stated is no.



Let $V = overlineoperatornamespan e_2n_ninmathbbN$ and $U = overlineoperatornamespan lefte_2n + frace_2n+1n+1right_ninmathbbN$.



Then $U, V$ are infinite-dimensional closed subspaces of $ell^2$ and $V cong V$, $V cong U$. The latter are isometrically isomorphic via $$e_2n mapsto frac1sqrt1+frac1(n+1)^2 left(e_2n + frace_2n+1n+1right)$$



However, it is shown here that $V + U$ is not closed so it is not complete. But $V$ is complete so $V$ and $V + U$ cannot be isometrically isomorphic.






share|cite|improve this answer





















  • thanks for your answer
    – Matey Math
    Jul 22 at 19:46












up vote
1
down vote



accepted







up vote
1
down vote



accepted






The answer to the question as stated is no.



Let $V = overlineoperatornamespan e_2n_ninmathbbN$ and $U = overlineoperatornamespan lefte_2n + frace_2n+1n+1right_ninmathbbN$.



Then $U, V$ are infinite-dimensional closed subspaces of $ell^2$ and $V cong V$, $V cong U$. The latter are isometrically isomorphic via $$e_2n mapsto frac1sqrt1+frac1(n+1)^2 left(e_2n + frace_2n+1n+1right)$$



However, it is shown here that $V + U$ is not closed so it is not complete. But $V$ is complete so $V$ and $V + U$ cannot be isometrically isomorphic.






share|cite|improve this answer













The answer to the question as stated is no.



Let $V = overlineoperatornamespan e_2n_ninmathbbN$ and $U = overlineoperatornamespan lefte_2n + frace_2n+1n+1right_ninmathbbN$.



Then $U, V$ are infinite-dimensional closed subspaces of $ell^2$ and $V cong V$, $V cong U$. The latter are isometrically isomorphic via $$e_2n mapsto frac1sqrt1+frac1(n+1)^2 left(e_2n + frace_2n+1n+1right)$$



However, it is shown here that $V + U$ is not closed so it is not complete. But $V$ is complete so $V$ and $V + U$ cannot be isometrically isomorphic.







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 22 at 19:43









mechanodroid

22.2k52041




22.2k52041











  • thanks for your answer
    – Matey Math
    Jul 22 at 19:46
















  • thanks for your answer
    – Matey Math
    Jul 22 at 19:46















thanks for your answer
– Matey Math
Jul 22 at 19:46




thanks for your answer
– Matey Math
Jul 22 at 19:46










up vote
1
down vote













No in general, as sum of two closed subspace is not necessarily closed. See: The direct sum of two closed subspace is closed? (Hilbert space)



However, if $W + U$ happens to be closed, then both $V$ and $W + U$ will be isomorphic to $ell^2$ itself, as infinite dimensional subspace of separable Hilbert space is also a separable Hilbert space. Hence, $V cong ell^2 cong W + U$.






share|cite|improve this answer























  • now I edited that the spaces are infinite dimensional.
    – Matey Math
    Jul 22 at 19:06











  • In fact every Hilbert space with countable Schauder Basis (or eqv. separable) is isomorphic to $ell^2$. This is true for closed $infty$-dimensional subspace of $ell^2$. So the answer must be yes now.
    – Nik Pronko
    Jul 22 at 19:18











  • thanks @Nik Pronko for your answer P.S. if you put your comment in an answer i can sign it as answered
    – Matey Math
    Jul 22 at 19:20











  • No it still can be wrong as sum of two closed subspaces is not necessarily closed. See: math.stackexchange.com/questions/135471/…
    – Nik Pronko
    Jul 22 at 19:28






  • 1




    Yes, then all subspaces just isomorphic to $ell^2$.
    – Nik Pronko
    Jul 22 at 19:34














up vote
1
down vote













No in general, as sum of two closed subspace is not necessarily closed. See: The direct sum of two closed subspace is closed? (Hilbert space)



However, if $W + U$ happens to be closed, then both $V$ and $W + U$ will be isomorphic to $ell^2$ itself, as infinite dimensional subspace of separable Hilbert space is also a separable Hilbert space. Hence, $V cong ell^2 cong W + U$.






share|cite|improve this answer























  • now I edited that the spaces are infinite dimensional.
    – Matey Math
    Jul 22 at 19:06











  • In fact every Hilbert space with countable Schauder Basis (or eqv. separable) is isomorphic to $ell^2$. This is true for closed $infty$-dimensional subspace of $ell^2$. So the answer must be yes now.
    – Nik Pronko
    Jul 22 at 19:18











  • thanks @Nik Pronko for your answer P.S. if you put your comment in an answer i can sign it as answered
    – Matey Math
    Jul 22 at 19:20











  • No it still can be wrong as sum of two closed subspaces is not necessarily closed. See: math.stackexchange.com/questions/135471/…
    – Nik Pronko
    Jul 22 at 19:28






  • 1




    Yes, then all subspaces just isomorphic to $ell^2$.
    – Nik Pronko
    Jul 22 at 19:34












up vote
1
down vote










up vote
1
down vote









No in general, as sum of two closed subspace is not necessarily closed. See: The direct sum of two closed subspace is closed? (Hilbert space)



However, if $W + U$ happens to be closed, then both $V$ and $W + U$ will be isomorphic to $ell^2$ itself, as infinite dimensional subspace of separable Hilbert space is also a separable Hilbert space. Hence, $V cong ell^2 cong W + U$.






share|cite|improve this answer















No in general, as sum of two closed subspace is not necessarily closed. See: The direct sum of two closed subspace is closed? (Hilbert space)



However, if $W + U$ happens to be closed, then both $V$ and $W + U$ will be isomorphic to $ell^2$ itself, as infinite dimensional subspace of separable Hilbert space is also a separable Hilbert space. Hence, $V cong ell^2 cong W + U$.







share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited Jul 22 at 19:43


























answered Jul 22 at 19:02









Nik Pronko

795717




795717











  • now I edited that the spaces are infinite dimensional.
    – Matey Math
    Jul 22 at 19:06











  • In fact every Hilbert space with countable Schauder Basis (or eqv. separable) is isomorphic to $ell^2$. This is true for closed $infty$-dimensional subspace of $ell^2$. So the answer must be yes now.
    – Nik Pronko
    Jul 22 at 19:18











  • thanks @Nik Pronko for your answer P.S. if you put your comment in an answer i can sign it as answered
    – Matey Math
    Jul 22 at 19:20











  • No it still can be wrong as sum of two closed subspaces is not necessarily closed. See: math.stackexchange.com/questions/135471/…
    – Nik Pronko
    Jul 22 at 19:28






  • 1




    Yes, then all subspaces just isomorphic to $ell^2$.
    – Nik Pronko
    Jul 22 at 19:34
















  • now I edited that the spaces are infinite dimensional.
    – Matey Math
    Jul 22 at 19:06











  • In fact every Hilbert space with countable Schauder Basis (or eqv. separable) is isomorphic to $ell^2$. This is true for closed $infty$-dimensional subspace of $ell^2$. So the answer must be yes now.
    – Nik Pronko
    Jul 22 at 19:18











  • thanks @Nik Pronko for your answer P.S. if you put your comment in an answer i can sign it as answered
    – Matey Math
    Jul 22 at 19:20











  • No it still can be wrong as sum of two closed subspaces is not necessarily closed. See: math.stackexchange.com/questions/135471/…
    – Nik Pronko
    Jul 22 at 19:28






  • 1




    Yes, then all subspaces just isomorphic to $ell^2$.
    – Nik Pronko
    Jul 22 at 19:34















now I edited that the spaces are infinite dimensional.
– Matey Math
Jul 22 at 19:06





now I edited that the spaces are infinite dimensional.
– Matey Math
Jul 22 at 19:06













In fact every Hilbert space with countable Schauder Basis (or eqv. separable) is isomorphic to $ell^2$. This is true for closed $infty$-dimensional subspace of $ell^2$. So the answer must be yes now.
– Nik Pronko
Jul 22 at 19:18





In fact every Hilbert space with countable Schauder Basis (or eqv. separable) is isomorphic to $ell^2$. This is true for closed $infty$-dimensional subspace of $ell^2$. So the answer must be yes now.
– Nik Pronko
Jul 22 at 19:18













thanks @Nik Pronko for your answer P.S. if you put your comment in an answer i can sign it as answered
– Matey Math
Jul 22 at 19:20





thanks @Nik Pronko for your answer P.S. if you put your comment in an answer i can sign it as answered
– Matey Math
Jul 22 at 19:20













No it still can be wrong as sum of two closed subspaces is not necessarily closed. See: math.stackexchange.com/questions/135471/…
– Nik Pronko
Jul 22 at 19:28




No it still can be wrong as sum of two closed subspaces is not necessarily closed. See: math.stackexchange.com/questions/135471/…
– Nik Pronko
Jul 22 at 19:28




1




1




Yes, then all subspaces just isomorphic to $ell^2$.
– Nik Pronko
Jul 22 at 19:34




Yes, then all subspaces just isomorphic to $ell^2$.
– Nik Pronko
Jul 22 at 19:34












 

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