Homeomorphism between a normed space and its open unit ball
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I have studied that every normed space $X$ is homeomorphic to its open unit ball $B$. I want to know what conclusion can we draw from this statement? Does it mean that every normed space is open ball? I am confused with this statement. Could anyone clear my doubt?
Thanks for help
functional-analysis
asked Jul 4 '12 at 4:28
srijansrijan
6,44063980
$endgroup$
add a comment |
$begingroup$
I have studied that every normed space $X$ is homeomorphic to its open unit ball $B$. I want to know what conclusion can we draw from this statement? Does it mean that every normed space is open ball? I am confused with this statement. Could anyone clear my doubt?
Thanks for help
functional-analysis
asked Jul 4 '12 at 4:28
srijansrijan
6,44063980
$endgroup$
$begingroup$
From @Alireza Imanzadeh fard: Not true! $f$ is not well defined. Even with the real field, $X$ can not be called a scalar product of its members? Is $X$ an integer with tg(π/2||x||).x|xinX? But this may be true f:X→B with f(x)=(1/(1+||x||).x ker(f)=0 and x=(1/(1−||y||)y f and f−1 cont.
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– dantopa
Jan 11 at 6:03
add a comment |
$begingroup$
I have studied that every normed space $X$ is homeomorphic to its open unit ball $B$. I want to know what conclusion can we draw from this statement? Does it mean that every normed space is open ball? I am confused with this statement. Could anyone clear my doubt?
Thanks for help
functional-analysis
asked Jul 4 '12 at 4:28
srijansrijan
6,44063980
$endgroup$
I have studied that every normed space $X$ is homeomorphic to its open unit ball $B$. I want to know what conclusion can we draw from this statement? Does it mean that every normed space is open ball? I am confused with this statement. Could anyone clear my doubt?
Thanks for help
functional-analysis
functional-analysis
asked Jul 4 '12 at 4:28
srijansrijan
6,44063980
asked Jul 4 '12 at 4:28
srijansrijan
6,44063980
edited Jul 4 '12 at 4:34
srijan
asked Jul 4 '12 at 4:28
srijansrijan
6,44063980
asked Jul 4 '12 at 4:28
srijansrijan
6,44063980
asked Jul 4 '12 at 4:28
srijansrijan
6,44063980
6,44063980
$begingroup$
From @Alireza Imanzadeh fard: Not true! $f$ is not well defined. Even with the real field, $X$ can not be called a scalar product of its members? Is $X$ an integer with tg(π/2||x||).x|xinX? But this may be true f:X→B with f(x)=(1/(1+||x||).x ker(f)=0 and x=(1/(1−||y||)y f and f−1 cont.
$endgroup$
– dantopa
Jan 11 at 6:03
add a comment |
$begingroup$
From @Alireza Imanzadeh fard: Not true! $f$ is not well defined. Even with the real field, $X$ can not be called a scalar product of its members? Is $X$ an integer with tg(π/2||x||).x|xinX? But this may be true f:X→B with f(x)=(1/(1+||x||).x ker(f)=0 and x=(1/(1−||y||)y f and f−1 cont.
$endgroup$
– dantopa
Jan 11 at 6:03
$begingroup$
From @Alireza Imanzadeh fard: Not true! $f$ is not well defined. Even with the real field, $X$ can not be called a scalar product of its members? Is $X$ an integer with tg(π/2||x||).x|xinX? But this may be true f:X→B with f(x)=(1/(1+||x||).x ker(f)=0 and x=(1/(1−||y||)y f and f−1 cont.
$endgroup$
– dantopa
Jan 11 at 6:03
$begingroup$
From @Alireza Imanzadeh fard: Not true! $f$ is not well defined. Even with the real field, $X$ can not be called a scalar product of its members? Is $X$ an integer with tg(π/2||x||).x|xinX? But this may be true f:X→B with f(x)=(1/(1+||x||).x ker(f)=0 and x=(1/(1−||y||)y f and f−1 cont.
$endgroup$
– dantopa
Jan 11 at 6:03
add a comment |
1 Answer
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It is true that every normed linear space $X$ is homeomorphic to its open unit ball $B$, for example via the homeomorphism $f:Bto X$ defined by $f(x)=tan(frac{pi}{2}|x|)cdot x$. This means that, as topological spaces, $X$ and $B$ are essentially identical. However, a normed linear space carries more information than a topological space, namely a norm. Thus the appropriate notion of "essentially identical" for normed linear space is the existence of an isometric isomorphism, a homeomorphism which preserves the norm. Clearly $X$ is not isometrically isomorphic to $B$, since $B$ is not even a normed linear space unless $X={0}$. To see this, take any $xneq 0$. Then $x/2|x|in B$ but $2cdot x/2|x|notin B$.
edited Jul 4 '12 at 14:26
Jonas Meyer
40.8k6148256
answered Jul 4 '12 at 5:03
Alex BeckerAlex Becker
49.1k698161
$endgroup$
2
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As a normed linear space, $B$ isn't.
$endgroup$
– Robert Israel
Jul 4 '12 at 5:43
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@RobertIsrael Indeed, I should probably mention that.
$endgroup$
– Alex Becker
Jul 4 '12 at 5:51
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$B$ can't have a norm because it's not a linear space, but it does have a metric. $X$ is not isometric to $B$ ($B$ is bounded, but $X$ is not).
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– Robert Israel
Jul 4 '12 at 17:02
add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
It is true that every normed linear space $X$ is homeomorphic to its open unit ball $B$, for example via the homeomorphism $f:Bto X$ defined by $f(x)=tan(frac{pi}{2}|x|)cdot x$. This means that, as topological spaces, $X$ and $B$ are essentially identical. However, a normed linear space carries more information than a topological space, namely a norm. Thus the appropriate notion of "essentially identical" for normed linear space is the existence of an isometric isomorphism, a homeomorphism which preserves the norm. Clearly $X$ is not isometrically isomorphic to $B$, since $B$ is not even a normed linear space unless $X={0}$. To see this, take any $xneq 0$. Then $x/2|x|in B$ but $2cdot x/2|x|notin B$.
edited Jul 4 '12 at 14:26
Jonas Meyer
40.8k6148256
answered Jul 4 '12 at 5:03
Alex BeckerAlex Becker
49.1k698161
$endgroup$
2
$begingroup$
As a normed linear space, $B$ isn't.
$endgroup$
– Robert Israel
Jul 4 '12 at 5:43
$begingroup$
@RobertIsrael Indeed, I should probably mention that.
$endgroup$
– Alex Becker
Jul 4 '12 at 5:51
$begingroup$
$B$ can't have a norm because it's not a linear space, but it does have a metric. $X$ is not isometric to $B$ ($B$ is bounded, but $X$ is not).
$endgroup$
– Robert Israel
Jul 4 '12 at 17:02
add a comment |
$begingroup$
It is true that every normed linear space $X$ is homeomorphic to its open unit ball $B$, for example via the homeomorphism $f:Bto X$ defined by $f(x)=tan(frac{pi}{2}|x|)cdot x$. This means that, as topological spaces, $X$ and $B$ are essentially identical. However, a normed linear space carries more information than a topological space, namely a norm. Thus the appropriate notion of "essentially identical" for normed linear space is the existence of an isometric isomorphism, a homeomorphism which preserves the norm. Clearly $X$ is not isometrically isomorphic to $B$, since $B$ is not even a normed linear space unless $X={0}$. To see this, take any $xneq 0$. Then $x/2|x|in B$ but $2cdot x/2|x|notin B$.
edited Jul 4 '12 at 14:26
Jonas Meyer
40.8k6148256
answered Jul 4 '12 at 5:03
Alex BeckerAlex Becker
49.1k698161
$endgroup$
2
$begingroup$
As a normed linear space, $B$ isn't.
$endgroup$
– Robert Israel
Jul 4 '12 at 5:43
$begingroup$
@RobertIsrael Indeed, I should probably mention that.
$endgroup$
– Alex Becker
Jul 4 '12 at 5:51
$begingroup$
$B$ can't have a norm because it's not a linear space, but it does have a metric. $X$ is not isometric to $B$ ($B$ is bounded, but $X$ is not).
$endgroup$
– Robert Israel
Jul 4 '12 at 17:02
add a comment |
$begingroup$
It is true that every normed linear space $X$ is homeomorphic to its open unit ball $B$, for example via the homeomorphism $f:Bto X$ defined by $f(x)=tan(frac{pi}{2}|x|)cdot x$. This means that, as topological spaces, $X$ and $B$ are essentially identical. However, a normed linear space carries more information than a topological space, namely a norm. Thus the appropriate notion of "essentially identical" for normed linear space is the existence of an isometric isomorphism, a homeomorphism which preserves the norm. Clearly $X$ is not isometrically isomorphic to $B$, since $B$ is not even a normed linear space unless $X={0}$. To see this, take any $xneq 0$. Then $x/2|x|in B$ but $2cdot x/2|x|notin B$.
edited Jul 4 '12 at 14:26
Jonas Meyer
40.8k6148256
answered Jul 4 '12 at 5:03
Alex BeckerAlex Becker
49.1k698161
$endgroup$
It is true that every normed linear space $X$ is homeomorphic to its open unit ball $B$, for example via the homeomorphism $f:Bto X$ defined by $f(x)=tan(frac{pi}{2}|x|)cdot x$. This means that, as topological spaces, $X$ and $B$ are essentially identical. However, a normed linear space carries more information than a topological space, namely a norm. Thus the appropriate notion of "essentially identical" for normed linear space is the existence of an isometric isomorphism, a homeomorphism which preserves the norm. Clearly $X$ is not isometrically isomorphic to $B$, since $B$ is not even a normed linear space unless $X={0}$. To see this, take any $xneq 0$. Then $x/2|x|in B$ but $2cdot x/2|x|notin B$.
edited Jul 4 '12 at 14:26
Jonas Meyer
40.8k6148256
answered Jul 4 '12 at 5:03
Alex BeckerAlex Becker
49.1k698161
edited Jul 4 '12 at 14:26
Jonas Meyer
40.8k6148256
edited Jul 4 '12 at 14:26
Jonas Meyer
40.8k6148256
edited Jul 4 '12 at 14:26
Jonas Meyer
40.8k6148256
40.8k6148256
answered Jul 4 '12 at 5:03
Alex BeckerAlex Becker
49.1k698161
answered Jul 4 '12 at 5:03
Alex BeckerAlex Becker
49.1k698161
answered Jul 4 '12 at 5:03
Alex BeckerAlex Becker
49.1k698161
49.1k698161
2
$begingroup$
As a normed linear space, $B$ isn't.
$endgroup$
– Robert Israel
Jul 4 '12 at 5:43
$begingroup$
@RobertIsrael Indeed, I should probably mention that.
$endgroup$
– Alex Becker
Jul 4 '12 at 5:51
$begingroup$
$B$ can't have a norm because it's not a linear space, but it does have a metric. $X$ is not isometric to $B$ ($B$ is bounded, but $X$ is not).
$endgroup$
– Robert Israel
Jul 4 '12 at 17:02
add a comment |
2
$begingroup$
As a normed linear space, $B$ isn't.
$endgroup$
– Robert Israel
Jul 4 '12 at 5:43
$begingroup$
@RobertIsrael Indeed, I should probably mention that.
$endgroup$
– Alex Becker
Jul 4 '12 at 5:51
$begingroup$
$B$ can't have a norm because it's not a linear space, but it does have a metric. $X$ is not isometric to $B$ ($B$ is bounded, but $X$ is not).
$endgroup$
– Robert Israel
Jul 4 '12 at 17:02
2
2
$begingroup$
As a normed linear space, $B$ isn't.
$endgroup$
– Robert Israel
Jul 4 '12 at 5:43
$begingroup$
As a normed linear space, $B$ isn't.
$endgroup$
– Robert Israel
Jul 4 '12 at 5:43
$begingroup$
@RobertIsrael Indeed, I should probably mention that.
$endgroup$
– Alex Becker
Jul 4 '12 at 5:51
$begingroup$
@RobertIsrael Indeed, I should probably mention that.
$endgroup$
– Alex Becker
Jul 4 '12 at 5:51
$begingroup$
$B$ can't have a norm because it's not a linear space, but it does have a metric. $X$ is not isometric to $B$ ($B$ is bounded, but $X$ is not).
$endgroup$
– Robert Israel
Jul 4 '12 at 17:02
$begingroup$
$B$ can't have a norm because it's not a linear space, but it does have a metric. $X$ is not isometric to $B$ ($B$ is bounded, but $X$ is not).
$endgroup$
– Robert Israel
Jul 4 '12 at 17:02
add a comment |
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$begingroup$
From @Alireza Imanzadeh fard: Not true! $f$ is not well defined. Even with the real field, $X$ can not be called a scalar product of its members? Is $X$ an integer with tg(π/2||x||).x|xinX? But this may be true f:X→B with f(x)=(1/(1+||x||).x ker(f)=0 and x=(1/(1−||y||)y f and f−1 cont.
$endgroup$
– dantopa
Jan 11 at 6:03