Span of a linearly independent subset of a hilbert space is a subspace iff finite

waddles
Messages
3
Reaction score
0

Homework Statement



Let S be a linearly independent subset of a Hilbert space. Prove that span(S) is a subspace, that is a linear manifold and a closed set, if and only if S is finite.

Homework Equations





The Attempt at a Solution



Assuming S is finite means that S is a closed set (a finite subset of a metric space is closed). I think that this will help to prove span(S) is a closed set but I am a bit stuck.
 
Physics news on Phys.org
Sorry I have no idea
 
Last edited:
waddles said:

Homework Statement



Let S be a linearly independent subset of a Hilbert space. Prove that span(S) is a subspace, that is a linear manifold and a closed set, if and only if S is finite.

Homework Equations




The Attempt at a Solution



Assuming S is finite means that S is a closed set (a finite subset of a metric space is closed). I think that this will help to prove span(S) is a closed set but I am a bit stuck.

span(S) is the set of all FINITE linear combinations of elements in S. To see how that would make a problem for the set being closed if S is infinite, define a convergent series that contains multiples of an infinite number of elements of S.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top