Mixing
In a complex metrizable topological vector space, $d(0,alpha x)neq |alpha|d(0,x), alpha in mathbb C.$
$begingroup$
Let $(X,tau)$ be a complex metrizable topological vector space with the metric $d$. Does the following hold:
$$d(0,alpha x)=d(0,x), forall alpha in mathbb C, |alpha|=1 ?$$
In general, the following holds: $$d(0,alpha x)neq |alpha|d(0,x), alpha in mathbb C.$$
functional-analysis metric-spaces topological-vector-spaces metrizability
asked Jan 20 at 10:56
InfinityInfinity
627514
$endgroup$
add a comment |
$begingroup$
Let $(X,tau)$ be a complex metrizable topological vector space with the metric $d$. Does the following hold:
$$d(0,alpha x)=d(0,x), forall alpha in mathbb C, |alpha|=1 ?$$
In general, the following holds: $$d(0,alpha x)neq |alpha|d(0,x), alpha in mathbb C.$$
functional-analysis metric-spaces topological-vector-spaces metrizability
asked Jan 20 at 10:56
InfinityInfinity
627514
$endgroup$
$begingroup$
Which examples did you check?
$endgroup$
– Did
Jan 20 at 11:05
add a comment |
$begingroup$
Let $(X,tau)$ be a complex metrizable topological vector space with the metric $d$. Does the following hold:
$$d(0,alpha x)=d(0,x), forall alpha in mathbb C, |alpha|=1 ?$$
In general, the following holds: $$d(0,alpha x)neq |alpha|d(0,x), alpha in mathbb C.$$
functional-analysis metric-spaces topological-vector-spaces metrizability
asked Jan 20 at 10:56
InfinityInfinity
627514
$endgroup$
Let $(X,tau)$ be a complex metrizable topological vector space with the metric $d$. Does the following hold:
$$d(0,alpha x)=d(0,x), forall alpha in mathbb C, |alpha|=1 ?$$
In general, the following holds: $$d(0,alpha x)neq |alpha|d(0,x), alpha in mathbb C.$$
functional-analysis metric-spaces topological-vector-spaces metrizability
functional-analysis metric-spaces topological-vector-spaces metrizability
asked Jan 20 at 10:56
InfinityInfinity
627514
asked Jan 20 at 10:56
InfinityInfinity
627514
asked Jan 20 at 10:56
InfinityInfinity
627514
asked Jan 20 at 10:56
InfinityInfinity
627514
asked Jan 20 at 10:56
InfinityInfinity
627514
627514
$begingroup$
Which examples did you check?
$endgroup$
– Did
Jan 20 at 11:05
add a comment |
$begingroup$
Which examples did you check?
$endgroup$
– Did
Jan 20 at 11:05
$begingroup$
Which examples did you check?
$endgroup$
– Did
Jan 20 at 11:05
$begingroup$
Which examples did you check?
$endgroup$
– Did
Jan 20 at 11:05
add a comment |
1 Answer
1
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Absolutely not. One can easily find counterexamples, but it seems more important to understand why one should not expect such an equality. "Metrizable topological vector space" just means that the topology generated by the metric satisfies certain conditions (namely making the vector space operations continuous). This leaves plenty of room to break your proposed equality without changing the topology.
For a concrete example take $mathbb{C}$ and let
$$phicolonmathbb{C}tomathbb{C},,zmapsto operatorname{Re}z+4 mathrm{i}operatorname{Im} z.$$
This map is clearly continuous with continuous inverse, thus
$$
dcolonmathbb{C}times mathbb{C}to [0,infty),,d(z,w)=|phi(z)-phi(w)|
$$
induces the Eucliean topology on $mathbb{C}$ (which is of course a vector space topology). However $d(0,2)=2$, while $d(0,2i)=8$.
answered Jan 21 at 10:24
MaoWaoMaoWao
3,603617
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Absolutely not. One can easily find counterexamples, but it seems more important to understand why one should not expect such an equality. "Metrizable topological vector space" just means that the topology generated by the metric satisfies certain conditions (namely making the vector space operations continuous). This leaves plenty of room to break your proposed equality without changing the topology.
For a concrete example take $mathbb{C}$ and let
$$phicolonmathbb{C}tomathbb{C},,zmapsto operatorname{Re}z+4 mathrm{i}operatorname{Im} z.$$
This map is clearly continuous with continuous inverse, thus
$$
dcolonmathbb{C}times mathbb{C}to [0,infty),,d(z,w)=|phi(z)-phi(w)|
$$
induces the Eucliean topology on $mathbb{C}$ (which is of course a vector space topology). However $d(0,2)=2$, while $d(0,2i)=8$.
answered Jan 21 at 10:24
MaoWaoMaoWao
3,603617
$endgroup$
add a comment |
$begingroup$
Absolutely not. One can easily find counterexamples, but it seems more important to understand why one should not expect such an equality. "Metrizable topological vector space" just means that the topology generated by the metric satisfies certain conditions (namely making the vector space operations continuous). This leaves plenty of room to break your proposed equality without changing the topology.
For a concrete example take $mathbb{C}$ and let
$$phicolonmathbb{C}tomathbb{C},,zmapsto operatorname{Re}z+4 mathrm{i}operatorname{Im} z.$$
This map is clearly continuous with continuous inverse, thus
$$
dcolonmathbb{C}times mathbb{C}to [0,infty),,d(z,w)=|phi(z)-phi(w)|
$$
induces the Eucliean topology on $mathbb{C}$ (which is of course a vector space topology). However $d(0,2)=2$, while $d(0,2i)=8$.
answered Jan 21 at 10:24
MaoWaoMaoWao
3,603617
$endgroup$
add a comment |
$begingroup$
Absolutely not. One can easily find counterexamples, but it seems more important to understand why one should not expect such an equality. "Metrizable topological vector space" just means that the topology generated by the metric satisfies certain conditions (namely making the vector space operations continuous). This leaves plenty of room to break your proposed equality without changing the topology.
For a concrete example take $mathbb{C}$ and let
$$phicolonmathbb{C}tomathbb{C},,zmapsto operatorname{Re}z+4 mathrm{i}operatorname{Im} z.$$
This map is clearly continuous with continuous inverse, thus
$$
dcolonmathbb{C}times mathbb{C}to [0,infty),,d(z,w)=|phi(z)-phi(w)|
$$
induces the Eucliean topology on $mathbb{C}$ (which is of course a vector space topology). However $d(0,2)=2$, while $d(0,2i)=8$.
answered Jan 21 at 10:24
MaoWaoMaoWao
3,603617
$endgroup$
Absolutely not. One can easily find counterexamples, but it seems more important to understand why one should not expect such an equality. "Metrizable topological vector space" just means that the topology generated by the metric satisfies certain conditions (namely making the vector space operations continuous). This leaves plenty of room to break your proposed equality without changing the topology.
For a concrete example take $mathbb{C}$ and let
$$phicolonmathbb{C}tomathbb{C},,zmapsto operatorname{Re}z+4 mathrm{i}operatorname{Im} z.$$
This map is clearly continuous with continuous inverse, thus
$$
dcolonmathbb{C}times mathbb{C}to [0,infty),,d(z,w)=|phi(z)-phi(w)|
$$
induces the Eucliean topology on $mathbb{C}$ (which is of course a vector space topology). However $d(0,2)=2$, while $d(0,2i)=8$.
answered Jan 21 at 10:24
MaoWaoMaoWao
3,603617
answered Jan 21 at 10:24
MaoWaoMaoWao
3,603617
answered Jan 21 at 10:24
MaoWaoMaoWao
3,603617
answered Jan 21 at 10:24
MaoWaoMaoWao
3,603617
3,603617
add a comment |
add a comment |
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$begingroup$
Which examples did you check?
$endgroup$
– Did
Jan 20 at 11:05