Does the set of all real vectors exist? What is its cardinality?
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I'm doing a linear algebra problem and need to prove that all vectors $vecv$ have some property $P(vecv)$ . To do this, I want to write the set of all vectors $vecv$ that have this property as $vecv:P(vecv)$.
This got me thinking that this set only exists (by set comprehension) if the set of all vectors exist. Does it? What is its cardinality. If it does not, how do I prove things about every real vector?
linear-algebra elementary-set-theory
edited Aug 29 at 14:36
Andrés E. Caicedo
63.4k7154238
asked Aug 29 at 4:42
Evan Rosica
504212
 |Â
show 5 more comments
up vote
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down vote
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I'm doing a linear algebra problem and need to prove that all vectors $vecv$ have some property $P(vecv)$ . To do this, I want to write the set of all vectors $vecv$ that have this property as $vecv:P(vecv)$.
This got me thinking that this set only exists (by set comprehension) if the set of all vectors exist. Does it? What is its cardinality. If it does not, how do I prove things about every real vector?
linear-algebra elementary-set-theory
edited Aug 29 at 14:36
Andrés E. Caicedo
63.4k7154238
asked Aug 29 at 4:42
Evan Rosica
504212
1
doing a linear algebra problemDoesn't that spell out the vector space it is working over?
– dxiv
Aug 29 at 4:45
It's a theoretical problem. I wanted to write the solution set of a given nonhomogeneous equation in terms of a solution to the corresponding homogeneous eqn as $vecp+vecv_h: vecv_h text solves Avecx=vec0 text and p text solves Avecx=vecb $ , but this got me wondering whether or not the set of all vectors, of which this is a subset is well defined. The problem does not discuss any dimension for $A$, but I assume it is finite.
– Evan Rosica
Aug 29 at 4:55
1
Finitely dimensional vector spaces certainly exist, you still appear to be assuming that $vecp, A$ etc are defined somehow. And if they didn't, any statement relying on their existence would be vacuously true, anyway.
– dxiv
Aug 29 at 4:59
2
Any singleton set $x$ can be made into a vector space by defining vector addition and scalar multiplication (in the only way possible—$x$ is the zero vector). That means, abstractly, that any object whatsoever is a vector in some vector space. In particular, the class of all vectors in the universe is far too large to be a set.
– Greg Martin
Aug 29 at 5:07
1
You can say "If $X$ is any vector-space over $Bbb R$ and if $v in X$ then...(etc)" without any reference to any possible set of all real vectors.
– DanielWainfleet
Aug 29 at 10:13
 |Â
show 5 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm doing a linear algebra problem and need to prove that all vectors $vecv$ have some property $P(vecv)$ . To do this, I want to write the set of all vectors $vecv$ that have this property as $vecv:P(vecv)$.
This got me thinking that this set only exists (by set comprehension) if the set of all vectors exist. Does it? What is its cardinality. If it does not, how do I prove things about every real vector?
linear-algebra elementary-set-theory
edited Aug 29 at 14:36
Andrés E. Caicedo
63.4k7154238
asked Aug 29 at 4:42
Evan Rosica
504212
I'm doing a linear algebra problem and need to prove that all vectors $vecv$ have some property $P(vecv)$ . To do this, I want to write the set of all vectors $vecv$ that have this property as $vecv:P(vecv)$.
This got me thinking that this set only exists (by set comprehension) if the set of all vectors exist. Does it? What is its cardinality. If it does not, how do I prove things about every real vector?
linear-algebra elementary-set-theory
edited Aug 29 at 14:36
Andrés E. Caicedo
63.4k7154238
asked Aug 29 at 4:42
Evan Rosica
504212
edited Aug 29 at 14:36
Andrés E. Caicedo
63.4k7154238
edited Aug 29 at 14:36
Andrés E. Caicedo
63.4k7154238
edited Aug 29 at 14:36
Andrés E. Caicedo
63.4k7154238
63.4k7154238
asked Aug 29 at 4:42
Evan Rosica
504212
asked Aug 29 at 4:42
Evan Rosica
504212
asked Aug 29 at 4:42
Evan Rosica
504212
504212
1
doing a linear algebra problemDoesn't that spell out the vector space it is working over?
– dxiv
Aug 29 at 4:45
It's a theoretical problem. I wanted to write the solution set of a given nonhomogeneous equation in terms of a solution to the corresponding homogeneous eqn as $vecp+vecv_h: vecv_h text solves Avecx=vec0 text and p text solves Avecx=vecb $ , but this got me wondering whether or not the set of all vectors, of which this is a subset is well defined. The problem does not discuss any dimension for $A$, but I assume it is finite.
– Evan Rosica
Aug 29 at 4:55
1
Finitely dimensional vector spaces certainly exist, you still appear to be assuming that $vecp, A$ etc are defined somehow. And if they didn't, any statement relying on their existence would be vacuously true, anyway.
– dxiv
Aug 29 at 4:59
2
Any singleton set $x$ can be made into a vector space by defining vector addition and scalar multiplication (in the only way possible—$x$ is the zero vector). That means, abstractly, that any object whatsoever is a vector in some vector space. In particular, the class of all vectors in the universe is far too large to be a set.
– Greg Martin
Aug 29 at 5:07
1
You can say "If $X$ is any vector-space over $Bbb R$ and if $v in X$ then...(etc)" without any reference to any possible set of all real vectors.
– DanielWainfleet
Aug 29 at 10:13
 |Â
show 5 more comments
1
doing a linear algebra problemDoesn't that spell out the vector space it is working over?
– dxiv
Aug 29 at 4:45
It's a theoretical problem. I wanted to write the solution set of a given nonhomogeneous equation in terms of a solution to the corresponding homogeneous eqn as $vecp+vecv_h: vecv_h text solves Avecx=vec0 text and p text solves Avecx=vecb $ , but this got me wondering whether or not the set of all vectors, of which this is a subset is well defined. The problem does not discuss any dimension for $A$, but I assume it is finite.
– Evan Rosica
Aug 29 at 4:55
1
Finitely dimensional vector spaces certainly exist, you still appear to be assuming that $vecp, A$ etc are defined somehow. And if they didn't, any statement relying on their existence would be vacuously true, anyway.
– dxiv
Aug 29 at 4:59
2
Any singleton set $x$ can be made into a vector space by defining vector addition and scalar multiplication (in the only way possible—$x$ is the zero vector). That means, abstractly, that any object whatsoever is a vector in some vector space. In particular, the class of all vectors in the universe is far too large to be a set.
– Greg Martin
Aug 29 at 5:07
1
You can say "If $X$ is any vector-space over $Bbb R$ and if $v in X$ then...(etc)" without any reference to any possible set of all real vectors.
– DanielWainfleet
Aug 29 at 10:13
1
1
doing a linear algebra problem Doesn't that spell out the vector space it is working over?– dxiv
Aug 29 at 4:45
doing a linear algebra problem Doesn't that spell out the vector space it is working over?– dxiv
Aug 29 at 4:45
It's a theoretical problem. I wanted to write the solution set of a given nonhomogeneous equation in terms of a solution to the corresponding homogeneous eqn as $vecp+vecv_h: vecv_h text solves Avecx=vec0 text and p text solves Avecx=vecb $ , but this got me wondering whether or not the set of all vectors, of which this is a subset is well defined. The problem does not discuss any dimension for $A$, but I assume it is finite.
– Evan Rosica
Aug 29 at 4:55
It's a theoretical problem. I wanted to write the solution set of a given nonhomogeneous equation in terms of a solution to the corresponding homogeneous eqn as $vecp+vecv_h: vecv_h text solves Avecx=vec0 text and p text solves Avecx=vecb $ , but this got me wondering whether or not the set of all vectors, of which this is a subset is well defined. The problem does not discuss any dimension for $A$, but I assume it is finite.
– Evan Rosica
Aug 29 at 4:55
1
1
Finitely dimensional vector spaces certainly exist, you still appear to be assuming that $vecp, A$ etc are defined somehow. And if they didn't, any statement relying on their existence would be vacuously true, anyway.
– dxiv
Aug 29 at 4:59
Finitely dimensional vector spaces certainly exist, you still appear to be assuming that $vecp, A$ etc are defined somehow. And if they didn't, any statement relying on their existence would be vacuously true, anyway.
– dxiv
Aug 29 at 4:59
2
2
Any singleton set $x$ can be made into a vector space by defining vector addition and scalar multiplication (in the only way possible—$x$ is the zero vector). That means, abstractly, that any object whatsoever is a vector in some vector space. In particular, the class of all vectors in the universe is far too large to be a set.
– Greg Martin
Aug 29 at 5:07
Any singleton set $x$ can be made into a vector space by defining vector addition and scalar multiplication (in the only way possible—$x$ is the zero vector). That means, abstractly, that any object whatsoever is a vector in some vector space. In particular, the class of all vectors in the universe is far too large to be a set.
– Greg Martin
Aug 29 at 5:07
1
1
You can say "If $X$ is any vector-space over $Bbb R$ and if $v in X$ then...(etc)" without any reference to any possible set of all real vectors.
– DanielWainfleet
Aug 29 at 10:13
You can say "If $X$ is any vector-space over $Bbb R$ and if $v in X$ then...(etc)" without any reference to any possible set of all real vectors.
– DanielWainfleet
Aug 29 at 10:13
 |Â
show 5 more comments
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1
doing a linear algebra problemDoesn't that spell out the vector space it is working over?– dxiv
Aug 29 at 4:45
It's a theoretical problem. I wanted to write the solution set of a given nonhomogeneous equation in terms of a solution to the corresponding homogeneous eqn as $vecp+vecv_h: vecv_h text solves Avecx=vec0 text and p text solves Avecx=vecb $ , but this got me wondering whether or not the set of all vectors, of which this is a subset is well defined. The problem does not discuss any dimension for $A$, but I assume it is finite.
– Evan Rosica
Aug 29 at 4:55
1
Finitely dimensional vector spaces certainly exist, you still appear to be assuming that $vecp, A$ etc are defined somehow. And if they didn't, any statement relying on their existence would be vacuously true, anyway.
– dxiv
Aug 29 at 4:59
2
Any singleton set $x$ can be made into a vector space by defining vector addition and scalar multiplication (in the only way possible—$x$ is the zero vector). That means, abstractly, that any object whatsoever is a vector in some vector space. In particular, the class of all vectors in the universe is far too large to be a set.
– Greg Martin
Aug 29 at 5:07
1
You can say "If $X$ is any vector-space over $Bbb R$ and if $v in X$ then...(etc)" without any reference to any possible set of all real vectors.
– DanielWainfleet
Aug 29 at 10:13