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Math Help - Convergence in sequences of sequences

  1. #1
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    Convergence in sequences of sequences

    Hi,

    I'm stuck on this problem:

    Let  (X, \| . \|_{\infty} ) be a normed vector space of complex sequences ( \alpha_{j} ) that converge to zero, with the sup-norm:

     \| ( \alpha_{j} ) \|_{\infty} = sup \{ | \alpha | : j \in \mathbb{N} \}

    Let  Z= \{ (\alpha_{j} ) \in X : \alpha_{2j}=j^{2} \alpha_{2j-1}, j \in \mathbb{N} \}

    I need to show that Z is closed in X, but I'm getting a bit confused about sequences of sequences... here's what I got so far:

    We will use the fact that Z is closed (in a metric space, as is a NVS) iff it contains all its limit points, i.e.

    we want to show that whenever  ( \alpha_{j} )_{n} \in Z is a sequence in Z, which converges to some  ( x_{j}) \in X then we have that  (x_{j}) \in Z .

    So, we take an  ( \alpha_{j} )_{n} \subset Z , which converges to some  ( x_{j}) \in X . By definition, this means that

     \forall \epsilon > 0, \exists N \in \mathbb{N} \ s. \ t. \ \forall n>N, \ we\ have\ that\ \| (\alpha_{j})_{n} - (x_{j}) \|_{\infty} < \epsilon

    But how do I use this to show that  (x_{j}) \in Z ?

    Possibly by contradiction - suppose  (x_{j}) is not in Z, then at least one of its members will not satisfy  x_{2j}=j^{2} x_{2j-1} , say, this is the k-th member; so we set  \epsilon = |x_{k}-k^{2} x_{2k-1}| and show that this contradicts convergence? But I'm still getting very confused in passing from the infinity-norm to the Euclidean norm...

    Is this right, where do I go from here, and is there any way to prove this without contradiction?
    Last edited by Mimi89; October 19th 2010 at 02:31 AM. Reason: wrote 'sequence element of Z' rather than 'sequence subset of Z'
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  2. #2
    Member Mauritzvdworm's Avatar
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    consider a sequence in Z, say \{(\alpha_j)_l\} which converges to some (\alpha_j)\in X, we can write
    \lim_{l\rightarrow \infty}(\alpha_j)_l=(a_j)\in X
    now, is this (\alpha_j) in Z?

    Consider the following
    \lim_{l\rightarrow \infty}(\alpha_{2j})_l=\lim_{l\rightarrow \infty}j^2(\alpha_{2j-1})_l
    so
    \alpha_{2j}=j^2(\alpha_{2j-1})
    and Z is closed
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  3. #3
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    Thanks a lot for that!

    One more question on the same problem:

    I have to show that the element

     x_{j}=(1,0,\frac{1}{4}, 0, \frac{1}{9}, o, ...) is in the closure of Y+Z but not in Y+Z.

    Now, I've written it as the limit of:

     ( 0, -1, 0, -1, 0, -1, ..., -1, 0, 0, 0, 0....) + <br />
(1, 1, \frac{1}{4}, 1, \frac{1}{9}, 1, ...., \frac{1}{n^{2}}, 1, 0, 0, 0, ...)

    where the former is in Y (the last '1' occurs at the 2n-th place) and the latter in Z (the last nonzero entry is in the 2n-th place, again). So, I've shown (xj) is in the closure, as I've written it as the limit of a sequence which is in the sum, and closures are closed.

    But how do I show it is *not* in the sum Y+Z?
    Last edited by Mimi89; October 19th 2010 at 08:28 AM.
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  4. #4
    Member Mauritzvdworm's Avatar
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    what is Y?
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  5. #5
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    Sorry - Y is the subset  \{ (\alpha_{j} ) \in X : \alpha_{2j-1} = 0, j \in \mathbb{N} \}

    And I've edited the above post; indeed, the first sequence in the sum does converge to 0 (just its limit doesn't).
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