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Math Help - precompactness in sequence spaces

  1. #1
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    precompactness in sequence spaces

    I have 2 questions about precompactness in sequence spacesDoh)
    1. Is A = {  (x_{n} : \sum_{n=1}^{\infty} \sqrt{n} |x_{n} | \leq 1 } is precompact in  c_{0} = set of all sequences converging to 0 equipped with the sup norm ?
    2. Is A= {  (x_{n} :  n x_{n} \rightarrow 0  as  n \rightarrow \infty } in  l_{1} ?
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  2. #2
    Super Member girdav's Avatar
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    1. We notice that if (x_n)_n\in A we have |x_n|\leq\frac 1{\sqrt n}\, \forall n. Given an \varepsilon>0, we can find a N such that \frac 1{\sqrt N}\leq \varepsilon. We can conclude because \left[0,1\right]^{N-1} is precompact.
    2. Let e^{(n)} the sequence defined by e^{(n)}_k=\begin{cases}1& \mbox{if } n=k\\<br />
0& \mbox{if } n\neq k<br />
\end{cases}.
    We notice that e^{(n)}\in A and if m\neq n,\: ||e^{(n)}-e^{(m)}||_{l^1}=\sqrt 2. We conclude that A is not precompact.
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