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If $F:Vto W$ is a continuous isomorphism, is it a homeomorphism?
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If $F:Vto W$ is a continuous isomorphism of vector spaces, is it a homeomorphism ?
I tried to prove that $F(U)$ is open when $U$ is open : Let $xin F(U)$, then $x=f(t)$ for a $tin U$. Since $U$ is open, there is a neighborhood $mathcal V(t)subset U$. I guess that $F(mathcal V(t))$ is an open include in $F(U)$ but I can't prove it.
If it's wrong, is the claim true if $V$ and $W$ are supposed to be norm space ? I tried to prove that $F^-1$ is continuous at $0$ : Suppose it's not. Then, there is a sequence $(x_n)$ of $V$ that converge to $0$ s.t. $$lim_nto infty F^-1(x_n)neq 0.$$
Let $y_n$ s.t. $x_n=F(y_n)$ (the sequence $(y_n)$ exist and is unique by bijectivity). We have that $lim_nto infty F(y_n)=0$.
Either $(y_n)$ is bounded, and there is a subsequence (still denoted $(y_n)$) that converge (to $ell$) and thus by continuity $F(ell)=0implies ell=0$ (because bijective) and thus it's a contradiction. Either it's unbounded, and thus I don't know.
- By the way, how would look a linear application that is not continuous (on $mathbb Rto mathbb R$ for example).
functional-analysis
asked Jul 24 at 13:42
user352653
354212
 |Â
show 5 more comments
up vote
1
down vote
favorite
If $F:Vto W$ is a continuous isomorphism of vector spaces, is it a homeomorphism ?
I tried to prove that $F(U)$ is open when $U$ is open : Let $xin F(U)$, then $x=f(t)$ for a $tin U$. Since $U$ is open, there is a neighborhood $mathcal V(t)subset U$. I guess that $F(mathcal V(t))$ is an open include in $F(U)$ but I can't prove it.
If it's wrong, is the claim true if $V$ and $W$ are supposed to be norm space ? I tried to prove that $F^-1$ is continuous at $0$ : Suppose it's not. Then, there is a sequence $(x_n)$ of $V$ that converge to $0$ s.t. $$lim_nto infty F^-1(x_n)neq 0.$$
Let $y_n$ s.t. $x_n=F(y_n)$ (the sequence $(y_n)$ exist and is unique by bijectivity). We have that $lim_nto infty F(y_n)=0$.
Either $(y_n)$ is bounded, and there is a subsequence (still denoted $(y_n)$) that converge (to $ell$) and thus by continuity $F(ell)=0implies ell=0$ (because bijective) and thus it's a contradiction. Either it's unbounded, and thus I don't know.
- By the way, how would look a linear application that is not continuous (on $mathbb Rto mathbb R$ for example).
functional-analysis
asked Jul 24 at 13:42
user352653
354212
2
What are $V$ and $W$? What kind of isomorphism is $F$? "Isomorphism" is very unspecific.
– M. Winter
Jul 24 at 13:46
@M.Winter: vector spaces.
– user352653
Jul 24 at 13:52
Do you assume that $V$ and $W$ are finite dimensional?
– M. Winter
Jul 24 at 13:54
@M.Winter: I don't. But let's say they are.
– user352653
Jul 24 at 13:56
1
If they are infinite-dimensional and Banach spaces, then the answer is yes by the open mapping theorem. In particular, the answer is yes in the finite-dimensinal case.
– noctusraid
Jul 24 at 13:58
 |Â
show 5 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
If $F:Vto W$ is a continuous isomorphism of vector spaces, is it a homeomorphism ?
I tried to prove that $F(U)$ is open when $U$ is open : Let $xin F(U)$, then $x=f(t)$ for a $tin U$. Since $U$ is open, there is a neighborhood $mathcal V(t)subset U$. I guess that $F(mathcal V(t))$ is an open include in $F(U)$ but I can't prove it.
If it's wrong, is the claim true if $V$ and $W$ are supposed to be norm space ? I tried to prove that $F^-1$ is continuous at $0$ : Suppose it's not. Then, there is a sequence $(x_n)$ of $V$ that converge to $0$ s.t. $$lim_nto infty F^-1(x_n)neq 0.$$
Let $y_n$ s.t. $x_n=F(y_n)$ (the sequence $(y_n)$ exist and is unique by bijectivity). We have that $lim_nto infty F(y_n)=0$.
Either $(y_n)$ is bounded, and there is a subsequence (still denoted $(y_n)$) that converge (to $ell$) and thus by continuity $F(ell)=0implies ell=0$ (because bijective) and thus it's a contradiction. Either it's unbounded, and thus I don't know.
- By the way, how would look a linear application that is not continuous (on $mathbb Rto mathbb R$ for example).
functional-analysis
asked Jul 24 at 13:42
user352653
354212
If $F:Vto W$ is a continuous isomorphism of vector spaces, is it a homeomorphism ?
I tried to prove that $F(U)$ is open when $U$ is open : Let $xin F(U)$, then $x=f(t)$ for a $tin U$. Since $U$ is open, there is a neighborhood $mathcal V(t)subset U$. I guess that $F(mathcal V(t))$ is an open include in $F(U)$ but I can't prove it.
If it's wrong, is the claim true if $V$ and $W$ are supposed to be norm space ? I tried to prove that $F^-1$ is continuous at $0$ : Suppose it's not. Then, there is a sequence $(x_n)$ of $V$ that converge to $0$ s.t. $$lim_nto infty F^-1(x_n)neq 0.$$
Let $y_n$ s.t. $x_n=F(y_n)$ (the sequence $(y_n)$ exist and is unique by bijectivity). We have that $lim_nto infty F(y_n)=0$.
Either $(y_n)$ is bounded, and there is a subsequence (still denoted $(y_n)$) that converge (to $ell$) and thus by continuity $F(ell)=0implies ell=0$ (because bijective) and thus it's a contradiction. Either it's unbounded, and thus I don't know.
- By the way, how would look a linear application that is not continuous (on $mathbb Rto mathbb R$ for example).
functional-analysis
asked Jul 24 at 13:42
user352653
354212
edited Jul 24 at 13:52
asked Jul 24 at 13:42
user352653
354212
asked Jul 24 at 13:42
user352653
354212
asked Jul 24 at 13:42
user352653
354212
354212
2
What are $V$ and $W$? What kind of isomorphism is $F$? "Isomorphism" is very unspecific.
– M. Winter
Jul 24 at 13:46
@M.Winter: vector spaces.
– user352653
Jul 24 at 13:52
Do you assume that $V$ and $W$ are finite dimensional?
– M. Winter
Jul 24 at 13:54
@M.Winter: I don't. But let's say they are.
– user352653
Jul 24 at 13:56
1
If they are infinite-dimensional and Banach spaces, then the answer is yes by the open mapping theorem. In particular, the answer is yes in the finite-dimensinal case.
– noctusraid
Jul 24 at 13:58
 |Â
show 5 more comments
2
What are $V$ and $W$? What kind of isomorphism is $F$? "Isomorphism" is very unspecific.
– M. Winter
Jul 24 at 13:46
@M.Winter: vector spaces.
– user352653
Jul 24 at 13:52
Do you assume that $V$ and $W$ are finite dimensional?
– M. Winter
Jul 24 at 13:54
@M.Winter: I don't. But let's say they are.
– user352653
Jul 24 at 13:56
1
If they are infinite-dimensional and Banach spaces, then the answer is yes by the open mapping theorem. In particular, the answer is yes in the finite-dimensinal case.
– noctusraid
Jul 24 at 13:58
2
2
What are $V$ and $W$? What kind of isomorphism is $F$? "Isomorphism" is very unspecific.
– M. Winter
Jul 24 at 13:46
What are $V$ and $W$? What kind of isomorphism is $F$? "Isomorphism" is very unspecific.
– M. Winter
Jul 24 at 13:46
@M.Winter: vector spaces.
– user352653
Jul 24 at 13:52
@M.Winter: vector spaces.
– user352653
Jul 24 at 13:52
Do you assume that $V$ and $W$ are finite dimensional?
– M. Winter
Jul 24 at 13:54
Do you assume that $V$ and $W$ are finite dimensional?
– M. Winter
Jul 24 at 13:54
@M.Winter: I don't. But let's say they are.
– user352653
Jul 24 at 13:56
@M.Winter: I don't. But let's say they are.
– user352653
Jul 24 at 13:56
1
1
If they are infinite-dimensional and Banach spaces, then the answer is yes by the open mapping theorem. In particular, the answer is yes in the finite-dimensinal case.
– noctusraid
Jul 24 at 13:58
If they are infinite-dimensional and Banach spaces, then the answer is yes by the open mapping theorem. In particular, the answer is yes in the finite-dimensinal case.
– noctusraid
Jul 24 at 13:58
 |Â
show 5 more comments
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
If $V, W$ are Banach spaces, than thanks to the open mapping theorem, a linear continuous surjective map between them is open; so, if it is also injective, it is an omeomorphism. Actually I don't know if this results holds in general for topological vector spaces that are not Banach, but in the proof of the Theorem the completeness is strongly used, so I think the answer is negative.
answered Jul 24 at 13:57
Giuseppe Bargagnati
984414
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
If $V, W$ are Banach spaces, than thanks to the open mapping theorem, a linear continuous surjective map between them is open; so, if it is also injective, it is an omeomorphism. Actually I don't know if this results holds in general for topological vector spaces that are not Banach, but in the proof of the Theorem the completeness is strongly used, so I think the answer is negative.
answered Jul 24 at 13:57
Giuseppe Bargagnati
984414
add a comment |Â
up vote
2
down vote
accepted
If $V, W$ are Banach spaces, than thanks to the open mapping theorem, a linear continuous surjective map between them is open; so, if it is also injective, it is an omeomorphism. Actually I don't know if this results holds in general for topological vector spaces that are not Banach, but in the proof of the Theorem the completeness is strongly used, so I think the answer is negative.
answered Jul 24 at 13:57
Giuseppe Bargagnati
984414
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
If $V, W$ are Banach spaces, than thanks to the open mapping theorem, a linear continuous surjective map between them is open; so, if it is also injective, it is an omeomorphism. Actually I don't know if this results holds in general for topological vector spaces that are not Banach, but in the proof of the Theorem the completeness is strongly used, so I think the answer is negative.
answered Jul 24 at 13:57
Giuseppe Bargagnati
984414
If $V, W$ are Banach spaces, than thanks to the open mapping theorem, a linear continuous surjective map between them is open; so, if it is also injective, it is an omeomorphism. Actually I don't know if this results holds in general for topological vector spaces that are not Banach, but in the proof of the Theorem the completeness is strongly used, so I think the answer is negative.
answered Jul 24 at 13:57
Giuseppe Bargagnati
984414
answered Jul 24 at 13:57
Giuseppe Bargagnati
984414
answered Jul 24 at 13:57
Giuseppe Bargagnati
984414
answered Jul 24 at 13:57
Giuseppe Bargagnati
984414
984414
add a comment |Â
add a comment |Â
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2
What are $V$ and $W$? What kind of isomorphism is $F$? "Isomorphism" is very unspecific.
– M. Winter
Jul 24 at 13:46
@M.Winter: vector spaces.
– user352653
Jul 24 at 13:52
Do you assume that $V$ and $W$ are finite dimensional?
– M. Winter
Jul 24 at 13:54
@M.Winter: I don't. But let's say they are.
– user352653
Jul 24 at 13:56
1
If they are infinite-dimensional and Banach spaces, then the answer is yes by the open mapping theorem. In particular, the answer is yes in the finite-dimensinal case.
– noctusraid
Jul 24 at 13:58