Comment on Listen here, Little Dicky
iAvicenna@lemmy.world 3 days agoDoesn’t BCT imply that an infinite dimensional Banach spaces cannot have a countable basis
Comment on Listen here, Little Dicky
iAvicenna@lemmy.world 3 days agoDoesn’t BCT imply that an infinite dimensional Banach spaces cannot have a countable basis
gandalf_der_12te@discuss.tchncs.de 2 days ago
Uhm, yeah, but there’s two different definitions of basis iirc. And i’m using the analytical definition here; you’re talking about the algebraic definition.
iAvicenna@lemmy.world 2 days ago
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?
gandalf_der_12te@discuss.tchncs.de 2 days ago
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.
iAvicenna@lemmy.world 2 days ago
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.
gandalf_der_12te@discuss.tchncs.de 2 days ago
i just checked and there’s official names for it: