Uhm, yeah, but there’s two different definitions of basis iirc. And i’m using the analytical definition here; you’re talking about the linear algebra definition.
So I call an infinite dimensional vector space of countable/uncountable dimensions if it has a countable and uncountable basis. What is the analytical definition? Or do you mean basis in the sense of topology?
Uhm, i remember there’s two definitions for basis.
The basis in linear algebra says that you can compose every vector v as a finite sum v = sum over i from 1 to N of a_i * v_i, where a_i are arbitrary coefficients
The basis in analysis says that you can compose every vector v as an infinite sum v = sum over i from 1 to infinity of a_i * v_i. So that makes a convergent series. It requires that a topology is defined on the vector space fist, so convergence becomes well-defined. We call such a vector space of countably infinite dimension if such a basis (v_1, v_2, …) exists that every vector v can be represented as a convergent series.
Ah that makes sense, regular definition of basis is not much of use in infinite dimension anyways as far as I recall. Wonder if differentiability is required for what you said since polynomials on compact domains (probably required for uniform convergence or sth) would also work for cont functions I think.
regular definition of basis is not much of use in infinite dimension anyways as far as I recall.
yeah, that’s exactly why we have an alternative definition for that :D
Wonder if differentiability is required for what you said since polynomials on compact domains (probably required for uniform convergence or sth) would also work for cont functions I think.
Differentiability is not required; what is required is a topology, i.e. a definition of convergence to make sure the infinite series are well-defined.
Look it is so simple, it just acts on an uncountably infinite dimensional vector space of differentiable functions.
fun fact: the vector space of differentiable functions (at least on compact domains) is actually of countable dimension.
still infinite though
Doesn’t BCT imply that infinite dimensional Banach spaces cannot have a countable basis
Uhm, yeah, but there’s two different definitions of basis iirc. And i’m using the analytical definition here; you’re talking about the linear algebra definition.
So I call an infinite dimensional vector space of countable/uncountable dimensions if it has a countable and uncountable basis. What is the analytical definition? Or do you mean basis in the sense of topology?
Uhm, i remember there’s two definitions for basis.
The basis in linear algebra says that you can compose every vector v as a finite sum v = sum over i from 1 to N of a_i * v_i, where a_i are arbitrary coefficients
The basis in analysis says that you can compose every vector v as an infinite sum v = sum over i from 1 to infinity of a_i * v_i. So that makes a convergent series. It requires that a topology is defined on the vector space fist, so convergence becomes well-defined. We call such a vector space of countably infinite dimension if such a basis (v_1, v_2, …) exists that every vector v can be represented as a convergent series.
Ah that makes sense, regular definition of basis is not much of use in infinite dimension anyways as far as I recall. Wonder if differentiability is required for what you said since polynomials on compact domains (probably required for uniform convergence or sth) would also work for cont functions I think.
yeah, that’s exactly why we have an alternative definition for that :D
Differentiability is not required; what is required is a topology, i.e. a definition of convergence to make sure the infinite series are well-defined.
i just checked and there’s official names for it: