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    Kind of a trivial question, but if you have the sequence space X=\{(x_{n})_{n\in\mathbb{N}}|x_{  n}\in\mathbb{F}\}, endowed with the metric

    \displaystyle d((x_{n}),(y_{n})):=\sum_{k \in \mathbb{N}}2^{-k}\frac{|x_{k}-y_{k}|}{1+|x_{k}-y_{k}|}

    How do you define d((x_{n}),(x_{m}))?
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    (Original post by RamocitoMorales)
    Kind of a trivial question, but if you have the sequence space X=\{(x_{n})_{n\in\mathbb{N}}|x_{  n}\in\mathbb{F}\}, endowed with the metric

    \displaystyle d((x_{n}),(y_{n})):=\sum_{k \in \mathbb{N}}2^{-k}\frac{|x_{k}-y_{k}|}{1+|x_{k}-y_{k}|}

    How do you define d((x_{n}),(x_{m}))?
    It's not clear what you're asking here: how are you taking the sequence (x_{n}) to differ from the sequence (x_{m})?
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    (Original post by Gregorius)
    It's not clear what you're asking here: how are you taking the sequence (x_{n}) to differ from the sequence (x_{m})?
    I define m,n\ge n_{0}\in\mathbb{N} for (x_{n})_{n\in\mathbb{N}}\in X a Cauchy sequence.
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    (Original post by RamocitoMorales)
    I define m,n\ge n_{0}\in\mathbb{N} for (x_{n})_{n\in\mathbb{N}}\in X a Cauchy sequence.
    So here you have m,n\ge n_{0}\in\mathbb{N} appearing to choose fixed constant integers m and n greater than some particular integer, but in the next (x_{n})_{n\in\mathbb{N}} you're using n as the dummy variable indexing your sequence.Confusing!

    Could you give an example to illustrate what you are asking?
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    (Original post by Gregorius)
    Could you give an example to illustrate what you are asking?
    The greater picture is that I'm trying to prove that the sequence space X=\{(x_{n})_{n\in\mathbb{N}}|x_{  n}\in\mathbb{F}\}, endowed with the metric

    \displaystyle d((x_{n}),(y_{n})):=\sum_{k \in \mathbb{N}}2^{-k}\frac{|x_{k}-y_{k}|}{1+|x_{k}-y_{k}|}

    is complete.

    A metric space is complete if all Cauchy sequences are convergent.

    I let (x_{n})_{n\in\mathbb{N}} be a Cauchy sequence, then \forall\epsilon>0 \exists n_{0}\in\mathbb{N} such that m,n\ge n_{0},

    d((x_{n}),(x_{m}))\le\epsilon

    I was having trouble defining d((x_{n}), (x_{m})). Is that more clear?

    Spoiler:
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    Fix natural numbers l,j if that makes it less confusing.
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    (Original post by RamocitoMorales)
    The greater picture is that I'm trying to prove that the sequence space X=\{(x_{n})_{n\in\mathbb{N}}|x_{  n}\in\mathbb{F}\}, endowed with the metric

    \displaystyle d((x_{n}),(y_{n})):=\sum_{k \in \mathbb{N}}2^{-k}\frac{|x_{k}-y_{k}|}{1+|x_{k}-y_{k}|}

    is complete.
    Right, I can see what you are trying to do now. The problem is that each "point" in the sequence space is a sequence of elements of your underlying field: \textbf{x} = (x_{n}). You need to consider a cauchy sequence of these "points", where each "point" is itself a sequence.\textbf{x}_k = (x_{n})_k. In other words, a sequence of sequences.
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    (Original post by Gregorius)
    Right, I can see what you are trying to do now. The problem is that each "point" in the sequence space is a sequence of elements of your underlying field: \textbf{x} = (x_{n}). You need to consider a cauchy sequence of these "points", where each "point" is itself a sequence.\textbf{x}_k = (x_{n})_k. In other words, a sequence of sequences.
    But then if we take \lim_{k\to\infty}x_{k}=\lim_{k \to\infty}\{x_{n}\}_{k}=x_{n} which is another sequence and not convergent to a point;
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    (Original post by RamocitoMorales)
    But then if we take \lim_{k\to\infty}x_{k}=\lim_{k \to\infty}\{x_{n}\}_{k}=x_{n} which is another sequence and not convergent to a point;
    Since X is a sequence space, a point in X is a sequence.

    More generally, in problems like this, it is desperately important that you are clear about notation, because you'll need to distinguish between sequences as opposed to sequences of sequences. That's why Gregorius was explictly using bold face for elements of X - they're elements, but they are still sequences (over \mathbb{F}.

    All your problems seem to be related to confusion about this.
 
 
 
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