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In my Metric Spaces course, I have come across sequence spaces and want to prove that $l^1 = \{(x_n): \sum_{n=1}^{\infty} |x_n| \text{converges}\}$ is complete.


Let $(x_n^{(1)}), (x_n^{(2)}), (x_n^{(3)}),...$ be a Cauchy sequence in $l^1$.

$\forall \epsilon>0,\ \exists N \text{ such that } \forall p,q \geq N, \ ||(x_n^{(p)})-(x_n^{(q)})||_1 = \sum_{n=1}^{\infty} |x_n^{(p)}-x_n^{(q)}| < \epsilon$.

This means that for any $n_0 \in \mathbb{N}$, we have $|x_{n_0}^{(p)}-x_{n_0}^{(q)}| < \epsilon, \ \ \forall p,q \geq N$.

$x_{n_0}^{(p)}$ is a real Cauchy sequence. Since real cauchy sequence converges, $x_{n_0}^{(p)} \rightarrow X_{n_0}$ as $p \rightarrow \infty$.

This is true for all $n \in \mathbb{N}$. So, let $(X_n)$ be a sequence $X_{n_0},X_{n_1},X_{n_2},...$

By letting $q \rightarrow \infty$ from above, we have

$\forall \epsilon>0,\ \exists N \text{ such that } \forall p \geq N, \ ||(x_n^{(p)})-(X_n)||_1 < \epsilon.$

So, $(x_n^{(p)}) $converges to $(X_n)$ as $p \rightarrow \infty$.


I am unsure on whether this is a correct, complete proof. Especially, the idea of 'sequence of sequence' confuses me. Any advice on how to improve this would be useful. Thank you.

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  • $\begingroup$ Cf. Kreyszig Erwin, Introductory Functional Analysis with Applications, Wiley, 1978, section 1.5-4 page 35. $\endgroup$ – Nosrati Nov 18 '15 at 12:09
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Your proof is neither complete not correct in your final statement. You got new sequence - $X_n$. Then you try to show that $\exists \lim_{p \to \infty}x_n^{(p)} = X_n$. You try to go by definition, showing that for any $\varepsilon>0$, we can find such $N = N(\varepsilon)$ that for $\forall p>N$, $\sum_{n=1}^{\infty}|x_n^{(p)}-X_n| < \varepsilon$. Taking the same $N(\varepsilon)$ as in your initial statement doesn't suffice, though: in general case, $\sum_{n=1}^{\infty}|x_n^{(p)}-x_n^{(q)}|$ may be less than $\sum_{n=1}^{\infty}|x_n^{(p)}-X_n|$. Hint: $|x_n^{(p)}-X_n| \le |x_n^{(p)}-x_n^{(q)}|+|x_n^{(q)}-X_n|$.

But even when you prove that $\exists \lim_{p \to \infty}x_n^{(p)} = X_n$, this doesn't state space completeness yet. You have to show that $X_n \in l^1$ as well.

Actually, most of such metric space statements get down to using triangle inequality several times with right additional points.

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