How can I fix my proof that $\ell^1$ is complete?

Question

In an exercise, my teacher asked us to prove that $\ell^1$ is a Banach space. I was able to do so, but there are two steps in my proof that I'm not quite so sure they are correct. This is what I came up with:

---

let $x^n=(x^n_1, x^n_2, ...)$ be a cauchy sequence of points in $\ell ^1$. Then, for all $\varepsilon>0$ exists a natural number $N$ such that:

$$n,m\geq N \implies\sum_{i=1}^\infty|x^n_i-x^m_i| \leq \varepsilon$$ $$\implies |x^n_i-x^m_i| \leq \varepsilon$$

So, for every $i$ the sequence $(x^n_i)_{n\in \mathbb N}$ is a Cauchy sequence on $\mathbb C$ and therefore it converges, let's call it's limit $y_i \in \mathbb C$.

We want to prove that $x^n \to y = (y_1,y_2,...)$

---

This is the first step of the proof that I think isn't correct. I tried but failed to prove that $y$ is indeed a point in $\ell^1$, this is: $$\sum_{i=1}^\infty |y_i|\leq\infty$$

---

To show that $\lim x^n = y$, let $\varepsilon>0$. Because $\lim_nx^n_i=y_i$, for all $i$ we have that there is a natural number $N_i$ such that: $$n\geq N_i \implies |x_i^n - y_i|\leq \varepsilon\frac{1}{2^i}$$

so $$\sum_{i=1}^k|x_i^n - y_i|\leq\varepsilon\sum_{i=1}^k \frac{1}{2^i}\leq \varepsilon$$ for $n \geq \max\{N_1,...,N_k\}.$

Because this is a bounded increasing sum, we have that it converges and $$||x^n - y|| = \lim_{k\to\infty}\sum_{i=1}^k|x_i^n - y_i|\leq \varepsilon$$

So $\lim x^n = y$

---

This last part is the one I'm most unsure of. While I had a finite sum, the condition $$\sum_{i=1}^k|x_i^n - y_i|\leq \varepsilon$$

Was true for $n \geq \max\{N_1,...,N_k\}$ but I don't know if this is the case when $k\to\infty$.

I thought about changing the definition of $n$ to $n \geq \sup\{N_1,...,N_k\}$ to accommodate the case where $k\to\infty$, but this solves nothing because either the set is unbounded, or it eventually stays the same at some point for large values of $k$.

How can I fix these two issues with my proof?

Answer 1

Pick $\epsilon>0$ and thus there exists $N$ such that $n,m\geq N\implies $ $||x_{n}-x_{m}||_{l^{1}}<\epsilon$ . So we have $\sum_{k=1}^{\infty}|x_{n}^{k}-x_{m}^{k}|<\epsilon$

Thus there also exists $L_{0}$ such that $\sum_{k=1}^{r}|x_{n}^{k}-x_{m}^{k}|<\epsilon\,,\forall \,r\geq L_{0}$ . In particular $\sum_{k=1}^{L_{0}}|x_{n}^{k}-x_{m}^{k}|<\epsilon$ .

So Pick $N_{0}$ large enough such that $|x_{n}^{k}-y^{k}|<\frac{\epsilon}{2^{k}}\,,k=1,2,...,L_{0} $ for all $n\geq\max(N_{0},N)=M_{0}(say)$ . Note that $N_{0}$ depends on $L_{0}$ which depends on $\epsilon$. Thus $N_{0}$ depends only on $\epsilon$.

Thus $|y^{k}-x_{n}^{k}|\leq |y^{k}-x^{k}_{M_{0}}|+|x^{k}_{M_{0}}-x^{k}_{n}|$

Thus we have for $n\geq \max(N,N_{0})$ $$\sum_{k=1}^{L_{0}}|y^{k}-x_{n}^{k}|<\epsilon\bigg(1-\frac{1}{2^{L_{0}}}\bigg)+\epsilon\leq 2\epsilon$$.

Above we have used that $\sum_{k=1}^{L_{0}}|x_{n}^{k}-x_{M_{0}}^{k}|<\epsilon$ on account of $x_{n}$ being Cauchy and $|y^{k}-x_{M_{0}}^{k}|<\frac{\epsilon}{2^{k}}\,\,,\forall k=1,2,...,L_{0}$

And that's basically what we want to prove . $||y-x_{n}||_{l^{1}}\to 0$ and the fact that $y\in l^{1}$ can be proven from the above by taking say $\epsilon=1$.

Answer 2

The last part is wrong essentially because of the reason that you stated. One way to proceed would be: Let $\epsilon>0$ be given. You know there exists $N$ so that for any $n\geq m\geq N$ such that $||x^n-x^m||<\epsilon/2$. Now if you consider for some $k\geq N$: \begin{align*} 0\leq \sum^k_{i=1} |x^n_i-y_i|, k \geq N \end{align*} Now for each of these components till the $k^{th}$ one, there are $N_i's>N$ so that $|x_i^{N_i} - y_i| < \epsilon/2k$. Now by triangle inequality, we have: \begin{align*} \sum^k_{i=1} |x^n_i-y_i|\leq \sum^k_{i=1} |x^n_i - x^{N_i}_i| + |x_i^{N_i} - y_i| <\epsilon \end{align*} Hence we have: \begin{align*} \sum^k_{i=1} |y_i| \leq ||x^N|| + \epsilon, \forall k \geq N \implies y \in l_1 \end{align*} and the previous part means $||x^n-y|| \rightarrow 0$ as $n\rightarrow \infty$