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Setting: Let $(x_n)$ be Cauchy in $\ell^2$ over $\mathbb{F} = \mathbb{C}$ or $\mathbb{R}$. I'm trying to show that $(x_n) \rightarrow x \in \ell^2$. That is, I'm trying to show that $\ell^2$ is complete in a particular way outlined below. I only used the first few steps of the proof because once I understand the third step I think I can understand how the rest of the proof unfolds.

Attempt:

  1. View $(x_k) = (x_1, x_2, \ldots)$ s.t.

    $$ x_k = x_{k,1}, x_{k,2}, \ldots $$

  2. For any $i \in \mathbb{N}$ we have that for large enough $n$ and $m$ that

    $$ \left| x_{i,m} - x_{i,n} \right| < \varepsilon $$

  3. Therefore

    $$ \sum_{i=1}^\infty \left| x_{i,m} - x_{i,n} \right| < \varepsilon $$

Question: Why is this last step true? I can see how if it is true, then for $k$ large enough we would have that $\{x_{i,k}\}_{i=1}^\infty \rightarrow y_i$ for some $y_i \in \mathbb{F}$ (since $\mathbb{F}$ is complete).

user1770201
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1 Answers1

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Notice that the large enough is independent of $i$, because we have $$\sum_{i=1}^{+\infty}|x_{i,m}-x_{i,n}|\lt\varepsilon $$ but it does not follow from what is mentioned in 2.

However, from what is given in 2., we can find, for each $i$, and $x_i$ such that $\lim_{n\to\infty}|x_{i,n}-x_i|=0$. The remaining part of the proof is to show that $x=(x_i)_{i\in\mathbb N}$ is in $\ell^2$ and that $\lVert x_n-x\rVert_2\to 0$ as $n$ goes to infinity.

Davide Giraudo
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