Recent content by ninty

I used instead the fact that exponentials are continuous Then E:={e^ax : a real } is a subset of C[a,b] It's clear that for finite n, {e^ix : i in N} is linearly independent. I'm fudging, but that would seem to imply that E is also linearly independent, since every finite subset is linearly...

Unless you are explicitly asked for a proof, a counterexample is enough

That is simply equivalent to the (generalized)definition of a basis. If you replace infinite-dimensional with finite, you'll get the usual definition. For infinite-dimensional cases this usual basis is called Hamel(Algebraic) basis. Whether such a basis exists depends on the assumption of...

We know that dim(C[a,b]) is infinte. Indeed it cannot be finite since it contains the set of all polynomials. Is the dimension of a Hamel basis for it countable or uncountable? I guess if we put a norm on it to make a Banach space, we could use Baire's to imply uncountable. I am however...

d) The set is an intersection of two sets. 0 for example is in only one set, but not the other, so it's not in the intersection. First think about what points are actually in the set, then try to figure out the interior.

Consider the definition of a homomorphism itself, then the elements on the image should form a group(which is a subgroup of G2) Try applying the subgroup test to P(G1), whichever one you've learned. My group theory is kinda rusty, so I'll leave the rest to others.

If you want to integrate term by term, you need uniform convergence. Haven't really looked at this, so not saying that term by term integration is the solution here.

2. Let A = the set of sequences with only finitely many non-zero components(N of them), where each term is a member of the rationals. We can show that the we can approximate every element of \ell^2 by sequences in A, hence the closure is \ell^2 . (The set \ell^2 \ A are the limit points)...

Is R connected? If so, this is just the maximum principle

By definition an eigenvector cannot be the zero vector. I'm assuming you're using the usual methods I see undergraduates use, (A-xI) = 0 If we represent vectors as (x1,x2,x3,x4,x5) we get x2=0, x5=0 Then solving for x1,x3,x4 we can get 3 linearly independent vectors which form a basis for...

How would continuity imply boundedness in this case? I would understand that if U was compact.

Could you not use polar coordinates instead? Then the equation would be much simpler and the limit is just as r goes to 0