Recent content by BSCowboy

Could someone please provide me a little direction. I'm trying to show this sequence converges x_n=n\left(e^{\frac{2\pi i}{n}}-1\right) I know \lim_{n\rightarrow\infty}|x_n|=2\pi i, but I have no idea how to arrive at the solution. This is not a homework question.

I know of l_p spaces. For 0\leq p<\infty it's the set whose elements are sequences of scalars x=\{\lambda_1, \lambda_2, \ldots, \lambda_n,\ldots\} such that \left(\sum|\lambda_n|^p\right)^{\frac{1}{p}} is convergent. But, I can't really help you with your problem.

Actually, I think I might have figured it out: f_n(t)=\frac{t^n}{n} \text{ in } \left(C[0,1],\|\cdot\|_1\right) \|f_n\|_1=\int_0^1f_n(t)dt=\frac{1}{n(n+1)} \text{ and } \sum_{n=1}^{\infty}\frac{1}{n(n+1)}\rightarrow 0 But, because the space \left(C[0,1],\|\cdot\|_1\right) is not...

I have been thinking about this problem: Determine whether the following series are convergent in \left(C[0,1],||\cdot ||_{\infty}\right) and \left(C[0,1],||\cdot ||_{1}\right). when f_n(t)=\frac{t^n}{n} In the supremum norm, this seems pretty straightforward, but in the integral norm I am...

I've been thinking about this some more; If B(x,1/2)\cap(E\setminus \{x\}) \Rightarrow ||y-x||_{\infty}<1/2\quad \text{and} \quad y\not=x This would be a contradiction since ||x-y||_{\infty}=1 \text{ if } x\not= y

Thanks, I appreciate your input.

Since (C[0,1],||\cdot||_{\infty}) is complete and \{f_n\} is convergent, we know every sequence in (C[0,1],||\cdot||_{\infty}) is Cauchy convergent and converges uniformly \Rightarrow \quad ||f_n-f||_{\infty}\rightarrow \, 0. Because of this we also know...

Yes, I think I've got it. As usual..it's seems pretty simple, now. Also, I appreciate you stubbornly sticking with me and my stubborn self.

Right, thank you. That is a good point. I haven't yet thought about this proclamation in all spaces. My statement was about the behavior was concerning this normed metric space.

You're right, I am having a hard time understanding sequences in E. Can you give me an example. See in E_o (the set of all sequences with a finite number of non-zero terms) an example would be: x_1=(1,0,\ldots) x_2=(1,\frac{1}{2},\ldots) x_n=(1,\frac{1}{2},\ldots, \frac{1}{n},\ldots) This...

Are you saying that every sequence in E is it's own limit and that since any sequence I construct in E will be it's own limit then E is complete?

That is my understanding

It would be the sequence containing only "c"?

In this case: A sequence x_n in \left(X,||\cdot||_{\infty}\right) is Cauchy iff \forall \epsilon >0\quad \exists N\quad \ni \quad ||x_n-x_m||_{\infty}<\epsilon \quad \forall m,n\geq N You are saying if we let \epsilon = 1/2 then: ||x_n-x_m||_{\infty}<1/2 \quad \Rightarrow \quad...

In this case: A sequence x_n in \left(X,||\cdot||_{\infty}\right) is Cauchy iff \forall \epsilon >0\quad \exists N\quad \ni \quad ||x_n-x_m||_{\infty}<\epsilon \quad \forall m,n\geq N You are saying if we let \epsilon = 1/2 then: ||x_n-x_m||_{\infty}<1/2 \quad \Rightarrow \quad...