Is the set of inner product spaces a subset of the set of metric spaces? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)What is it about modern set theory that prevents us from defining the set of all sets which are not members of themselves?Metric Space, Normed Space, and Inner Product space hierarcyConcepts of isomorphisms of linear spaces with a norm and inner productAssociation of a vector space to metric, normed and inner product spacesAre all the finite dimensional vector spaces with a metric isometric to $mathbb R^n$Generalization of inner product spaces (analogue to uniform spaces/locally convex spaces)Topological space $nRightarrow$ Metric space $nRightarrow$ Normed space $nRightarrow$ Inner product space (Examples)Metric spaces and normed vector spacesWhat is more general than a topological space?Is Hilbert space a Normed Space or a Inner Product Space? Or it have to be both at the same time?Relation between metric spaces, normed vector spaces, and inner product space.
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Is the set of inner product spaces a subset of the set of metric spaces?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)What is it about modern set theory that prevents us from defining the set of all sets which are not members of themselves?Metric Space, Normed Space, and Inner Product space hierarcyConcepts of isomorphisms of linear spaces with a norm and inner productAssociation of a vector space to metric, normed and inner product spacesAre all the finite dimensional vector spaces with a metric isometric to $mathbb R^n$Generalization of inner product spaces (analogue to uniform spaces/locally convex spaces)Topological space $nRightarrow$ Metric space $nRightarrow$ Normed space $nRightarrow$ Inner product space (Examples)Metric spaces and normed vector spacesWhat is more general than a topological space?Is Hilbert space a Normed Space or a Inner Product Space? Or it have to be both at the same time?Relation between metric spaces, normed vector spaces, and inner product space.
$begingroup$
I found this picture when looking up topological spaces.
Is this picture actually supposed to be interpreted as decreasing sets? That is, all inner product spaces are normed vector spaces, all metric spaces are topological spaces etc?
But this would mean that every inner product space and normed vector space was a metric space. However, I don't recall inner product spaces and normed vector spaces having a metric, so it's not a metric space. Right?
On the other hand Inner product spaces do have norms, so it is a normed space. So it does make sense that it's a subset of Normed vector space.
real-analysis general-topology analysis metric-spaces normed-spaces
edited Apr 2 at 18:14
José Carlos Santos
176k24137247
asked Apr 2 at 15:26
QwertfordQwertford
333212
$endgroup$
add a comment |
$begingroup$
I found this picture when looking up topological spaces.
Is this picture actually supposed to be interpreted as decreasing sets? That is, all inner product spaces are normed vector spaces, all metric spaces are topological spaces etc?
But this would mean that every inner product space and normed vector space was a metric space. However, I don't recall inner product spaces and normed vector spaces having a metric, so it's not a metric space. Right?
On the other hand Inner product spaces do have norms, so it is a normed space. So it does make sense that it's a subset of Normed vector space.
real-analysis general-topology analysis metric-spaces normed-spaces
edited Apr 2 at 18:14
José Carlos Santos
176k24137247
asked Apr 2 at 15:26
QwertfordQwertford
333212
$endgroup$
3
$begingroup$
In a normed vector space $(V,|cdot|)$ the metric is $d(x,y)=|x-y|.$
$endgroup$
– Thomas Andrews
Apr 2 at 15:28
6
$begingroup$
Right picture but none of these is a set.
$endgroup$
– Moishe Kohan
Apr 2 at 15:29
$begingroup$
Call these "classes" or "categories" or "universes" but do not call them "sets". math.stackexchange.com/questions/182618/…
$endgroup$
– Moishe Kohan
Apr 2 at 18:25
add a comment |
$begingroup$
I found this picture when looking up topological spaces.
Is this picture actually supposed to be interpreted as decreasing sets? That is, all inner product spaces are normed vector spaces, all metric spaces are topological spaces etc?
But this would mean that every inner product space and normed vector space was a metric space. However, I don't recall inner product spaces and normed vector spaces having a metric, so it's not a metric space. Right?
On the other hand Inner product spaces do have norms, so it is a normed space. So it does make sense that it's a subset of Normed vector space.
real-analysis general-topology analysis metric-spaces normed-spaces
edited Apr 2 at 18:14
José Carlos Santos
176k24137247
asked Apr 2 at 15:26
QwertfordQwertford
333212
$endgroup$
I found this picture when looking up topological spaces.
Is this picture actually supposed to be interpreted as decreasing sets? That is, all inner product spaces are normed vector spaces, all metric spaces are topological spaces etc?
But this would mean that every inner product space and normed vector space was a metric space. However, I don't recall inner product spaces and normed vector spaces having a metric, so it's not a metric space. Right?
On the other hand Inner product spaces do have norms, so it is a normed space. So it does make sense that it's a subset of Normed vector space.
real-analysis general-topology analysis metric-spaces normed-spaces
real-analysis general-topology analysis metric-spaces normed-spaces
edited Apr 2 at 18:14
José Carlos Santos
176k24137247
asked Apr 2 at 15:26
QwertfordQwertford
333212
edited Apr 2 at 18:14
José Carlos Santos
176k24137247
asked Apr 2 at 15:26
QwertfordQwertford
333212
edited Apr 2 at 18:14
José Carlos Santos
176k24137247
edited Apr 2 at 18:14
José Carlos Santos
176k24137247
edited Apr 2 at 18:14
José Carlos Santos
176k24137247
176k24137247
asked Apr 2 at 15:26
QwertfordQwertford
333212
asked Apr 2 at 15:26
QwertfordQwertford
333212
asked Apr 2 at 15:26
QwertfordQwertford
333212
333212
3
$begingroup$
In a normed vector space $(V,|cdot|)$ the metric is $d(x,y)=|x-y|.$
$endgroup$
– Thomas Andrews
Apr 2 at 15:28
6
$begingroup$
Right picture but none of these is a set.
$endgroup$
– Moishe Kohan
Apr 2 at 15:29
$begingroup$
Call these "classes" or "categories" or "universes" but do not call them "sets". math.stackexchange.com/questions/182618/…
$endgroup$
– Moishe Kohan
Apr 2 at 18:25
add a comment |
3
$begingroup$
In a normed vector space $(V,|cdot|)$ the metric is $d(x,y)=|x-y|.$
$endgroup$
– Thomas Andrews
Apr 2 at 15:28
6
$begingroup$
Right picture but none of these is a set.
$endgroup$
– Moishe Kohan
Apr 2 at 15:29
$begingroup$
Call these "classes" or "categories" or "universes" but do not call them "sets". math.stackexchange.com/questions/182618/…
$endgroup$
– Moishe Kohan
Apr 2 at 18:25
3
3
$begingroup$
In a normed vector space $(V,|cdot|)$ the metric is $d(x,y)=|x-y|.$
$endgroup$
– Thomas Andrews
Apr 2 at 15:28
$begingroup$
In a normed vector space $(V,|cdot|)$ the metric is $d(x,y)=|x-y|.$
$endgroup$
– Thomas Andrews
Apr 2 at 15:28
6
6
$begingroup$
Right picture but none of these is a set.
$endgroup$
– Moishe Kohan
Apr 2 at 15:29
$begingroup$
Right picture but none of these is a set.
$endgroup$
– Moishe Kohan
Apr 2 at 15:29
$begingroup$
Call these "classes" or "categories" or "universes" but do not call them "sets". math.stackexchange.com/questions/182618/…
$endgroup$
– Moishe Kohan
Apr 2 at 18:25
$begingroup$
Call these "classes" or "categories" or "universes" but do not call them "sets". math.stackexchange.com/questions/182618/…
$endgroup$
– Moishe Kohan
Apr 2 at 18:25
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If a vector space has an inner product, it has corresponding norm ($|x|=sqrtlangle x,xrangle$). Not all normed spaces are of this form (hence the proper inclusion)
If a vector space has a norm, it has a corresponding metric $d(x,y)=|x-y|$. Not all metric spaces are of this form (e.g. the discrete metric on $mathbbR$ is not, and moreover not all metric spaces have a natural underlying vector space structure that is needed for a norm).
If a set has a metric, it has a topology (the smallest one that makes all balls $B(x,r)$ open sets), but many topological spaces do not come from any metric (though there are theorems that tell us exactly when this is the case).
So the "inclusions" tell us that some structures naturally give rise to another more widely applicable structure.
answered Apr 2 at 17:11
Henno BrandsmaHenno Brandsma
117k350128
$endgroup$
add a comment |
$begingroup$
If $(V,lVertcdotrVert)$ is a normed space, then $V$ becomes a amtric space if we consider the distance $d$ defined as $d(v,w)=lVert v-wrVert$.
And not all norms come from inner products. Consider, for instance, in $mathbb R^n$ the norm$$bigllVert(a_1,ldots,a_n)bigrrVert_1=lvert a_1rvert+lvert a_2rvert+cdots+lvert a_nrvert.$$
answered Apr 2 at 15:32
José Carlos SantosJosé Carlos Santos
176k24137247
$endgroup$
$begingroup$
Well if we can just define a metric out of nothing, can I also say all metric spaces are inner product spaces because I can just define an inner product?
$endgroup$
– Qwertford
Apr 2 at 15:36
$begingroup$
How do you defined an inner product starting frome the discrete metric?
$endgroup$
– José Carlos Santos
Apr 2 at 15:37
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If a vector space has an inner product, it has corresponding norm ($|x|=sqrtlangle x,xrangle$). Not all normed spaces are of this form (hence the proper inclusion)
If a vector space has a norm, it has a corresponding metric $d(x,y)=|x-y|$. Not all metric spaces are of this form (e.g. the discrete metric on $mathbbR$ is not, and moreover not all metric spaces have a natural underlying vector space structure that is needed for a norm).
If a set has a metric, it has a topology (the smallest one that makes all balls $B(x,r)$ open sets), but many topological spaces do not come from any metric (though there are theorems that tell us exactly when this is the case).
So the "inclusions" tell us that some structures naturally give rise to another more widely applicable structure.
answered Apr 2 at 17:11
Henno BrandsmaHenno Brandsma
117k350128
$endgroup$
add a comment |
$begingroup$
If a vector space has an inner product, it has corresponding norm ($|x|=sqrtlangle x,xrangle$). Not all normed spaces are of this form (hence the proper inclusion)
If a vector space has a norm, it has a corresponding metric $d(x,y)=|x-y|$. Not all metric spaces are of this form (e.g. the discrete metric on $mathbbR$ is not, and moreover not all metric spaces have a natural underlying vector space structure that is needed for a norm).
If a set has a metric, it has a topology (the smallest one that makes all balls $B(x,r)$ open sets), but many topological spaces do not come from any metric (though there are theorems that tell us exactly when this is the case).
So the "inclusions" tell us that some structures naturally give rise to another more widely applicable structure.
answered Apr 2 at 17:11
Henno BrandsmaHenno Brandsma
117k350128
$endgroup$
add a comment |
$begingroup$
If a vector space has an inner product, it has corresponding norm ($|x|=sqrtlangle x,xrangle$). Not all normed spaces are of this form (hence the proper inclusion)
If a vector space has a norm, it has a corresponding metric $d(x,y)=|x-y|$. Not all metric spaces are of this form (e.g. the discrete metric on $mathbbR$ is not, and moreover not all metric spaces have a natural underlying vector space structure that is needed for a norm).
If a set has a metric, it has a topology (the smallest one that makes all balls $B(x,r)$ open sets), but many topological spaces do not come from any metric (though there are theorems that tell us exactly when this is the case).
So the "inclusions" tell us that some structures naturally give rise to another more widely applicable structure.
answered Apr 2 at 17:11
Henno BrandsmaHenno Brandsma
117k350128
$endgroup$
If a vector space has an inner product, it has corresponding norm ($|x|=sqrtlangle x,xrangle$). Not all normed spaces are of this form (hence the proper inclusion)
If a vector space has a norm, it has a corresponding metric $d(x,y)=|x-y|$. Not all metric spaces are of this form (e.g. the discrete metric on $mathbbR$ is not, and moreover not all metric spaces have a natural underlying vector space structure that is needed for a norm).
If a set has a metric, it has a topology (the smallest one that makes all balls $B(x,r)$ open sets), but many topological spaces do not come from any metric (though there are theorems that tell us exactly when this is the case).
So the "inclusions" tell us that some structures naturally give rise to another more widely applicable structure.
answered Apr 2 at 17:11
Henno BrandsmaHenno Brandsma
117k350128
answered Apr 2 at 17:11
Henno BrandsmaHenno Brandsma
117k350128
answered Apr 2 at 17:11
Henno BrandsmaHenno Brandsma
117k350128
answered Apr 2 at 17:11
Henno BrandsmaHenno Brandsma
117k350128
117k350128
add a comment |
add a comment |
$begingroup$
If $(V,lVertcdotrVert)$ is a normed space, then $V$ becomes a amtric space if we consider the distance $d$ defined as $d(v,w)=lVert v-wrVert$.
And not all norms come from inner products. Consider, for instance, in $mathbb R^n$ the norm$$bigllVert(a_1,ldots,a_n)bigrrVert_1=lvert a_1rvert+lvert a_2rvert+cdots+lvert a_nrvert.$$
answered Apr 2 at 15:32
José Carlos SantosJosé Carlos Santos
176k24137247
$endgroup$
$begingroup$
Well if we can just define a metric out of nothing, can I also say all metric spaces are inner product spaces because I can just define an inner product?
$endgroup$
– Qwertford
Apr 2 at 15:36
$begingroup$
How do you defined an inner product starting frome the discrete metric?
$endgroup$
– José Carlos Santos
Apr 2 at 15:37
add a comment |
$begingroup$
If $(V,lVertcdotrVert)$ is a normed space, then $V$ becomes a amtric space if we consider the distance $d$ defined as $d(v,w)=lVert v-wrVert$.
And not all norms come from inner products. Consider, for instance, in $mathbb R^n$ the norm$$bigllVert(a_1,ldots,a_n)bigrrVert_1=lvert a_1rvert+lvert a_2rvert+cdots+lvert a_nrvert.$$
answered Apr 2 at 15:32
José Carlos SantosJosé Carlos Santos
176k24137247
$endgroup$
$begingroup$
Well if we can just define a metric out of nothing, can I also say all metric spaces are inner product spaces because I can just define an inner product?
$endgroup$
– Qwertford
Apr 2 at 15:36
$begingroup$
How do you defined an inner product starting frome the discrete metric?
$endgroup$
– José Carlos Santos
Apr 2 at 15:37
add a comment |
$begingroup$
If $(V,lVertcdotrVert)$ is a normed space, then $V$ becomes a amtric space if we consider the distance $d$ defined as $d(v,w)=lVert v-wrVert$.
And not all norms come from inner products. Consider, for instance, in $mathbb R^n$ the norm$$bigllVert(a_1,ldots,a_n)bigrrVert_1=lvert a_1rvert+lvert a_2rvert+cdots+lvert a_nrvert.$$
answered Apr 2 at 15:32
José Carlos SantosJosé Carlos Santos
176k24137247
$endgroup$
If $(V,lVertcdotrVert)$ is a normed space, then $V$ becomes a amtric space if we consider the distance $d$ defined as $d(v,w)=lVert v-wrVert$.
And not all norms come from inner products. Consider, for instance, in $mathbb R^n$ the norm$$bigllVert(a_1,ldots,a_n)bigrrVert_1=lvert a_1rvert+lvert a_2rvert+cdots+lvert a_nrvert.$$
answered Apr 2 at 15:32
José Carlos SantosJosé Carlos Santos
176k24137247
answered Apr 2 at 15:32
José Carlos SantosJosé Carlos Santos
176k24137247
answered Apr 2 at 15:32
José Carlos SantosJosé Carlos Santos
176k24137247
answered Apr 2 at 15:32
José Carlos SantosJosé Carlos Santos
176k24137247
176k24137247
$begingroup$
Well if we can just define a metric out of nothing, can I also say all metric spaces are inner product spaces because I can just define an inner product?
$endgroup$
– Qwertford
Apr 2 at 15:36
$begingroup$
How do you defined an inner product starting frome the discrete metric?
$endgroup$
– José Carlos Santos
Apr 2 at 15:37
add a comment |
$begingroup$
Well if we can just define a metric out of nothing, can I also say all metric spaces are inner product spaces because I can just define an inner product?
$endgroup$
– Qwertford
Apr 2 at 15:36
$begingroup$
How do you defined an inner product starting frome the discrete metric?
$endgroup$
– José Carlos Santos
Apr 2 at 15:37
$begingroup$
Well if we can just define a metric out of nothing, can I also say all metric spaces are inner product spaces because I can just define an inner product?
$endgroup$
– Qwertford
Apr 2 at 15:36
$begingroup$
Well if we can just define a metric out of nothing, can I also say all metric spaces are inner product spaces because I can just define an inner product?
$endgroup$
– Qwertford
Apr 2 at 15:36
$begingroup$
How do you defined an inner product starting frome the discrete metric?
$endgroup$
– José Carlos Santos
Apr 2 at 15:37
$begingroup$
How do you defined an inner product starting frome the discrete metric?
$endgroup$
– José Carlos Santos
Apr 2 at 15:37
add a comment |
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$begingroup$
In a normed vector space $(V,|cdot|)$ the metric is $d(x,y)=|x-y|.$
$endgroup$
– Thomas Andrews
Apr 2 at 15:28
6
$begingroup$
Right picture but none of these is a set.
$endgroup$
– Moishe Kohan
Apr 2 at 15:29
$begingroup$
Call these "classes" or "categories" or "universes" but do not call them "sets". math.stackexchange.com/questions/182618/…
$endgroup$
– Moishe Kohan
Apr 2 at 18:25