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.

  • $\begingroup$ Cf. Kreyszig Erwin, Introductory Functional Analysis with Applications, Wiley, 1978, section 1.5-4 page 35. $\endgroup$ – Nosrati Nov 18 '15 at 12:09

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.