Proof confirmation that $\ell^2$ is complete Here's a proof I've written that $\ell^2$ is complete, I'd mostly like to know if there are any outstanding details I've missed (or notational nuances I might have screwed up along the way.)
Let $X^i$ be an $\ell^2$-Cauchy sequence. Then by definition we have that for all $\epsilon>0$ there exists an $N_\epsilon\in\mathbb{N}$ such that, for all $n,m>N_\epsilon$ we have $$||X^n-X^m||<\epsilon.$$ This means that for each $x^n_i,x^m_i$ that $(x^n_i-x^m_i)^2<\epsilon$ and since all norms on $\mathbb{R}$ are equivalent, then for each fixed i, $(x_i^n)$ converges to some $x_i$. Now we have that for each $\epsilon_i>0$ that $(x_i^n-x_i)^2<\epsilon_i$, for $n$ greater than some $N_i$. We can take this sequence to be such that $\epsilon>\sum_{i=1}^\infty \epsilon_i$, since each term, $(x_i^n-x_i)^2<\epsilon_i$, then for all $k$ we have $\sum_{i=1}^k(x_i^n-x_i)^2<\sum_{i=1}^k \epsilon_i<\epsilon$ which means $\sum_{i=1}^\infty (x_i^n-x_i)^2<\epsilon$ and hence $X^i\to X$.
Edit:
I thought my original proof was beyond repair, so wrote a different one that's a little more straight forwards.
Let $X^i$ be an $\ell^2$-Cauchy sequence in $\ell^2$ such that $X^n=(x^{(n)}_i)$. We have, for all $\epsilon>0$ there exists an $N_\epsilon\in\mathbb{N}$ such that for all $n,m>N_\epsilon$ we have \begin{equation}||X^n-X^m||=\sum_{i=1}^{\infty}(x_i^{(n)}-x_i^{(m)})^2<\epsilon\tag{1}\end{equation} This means that for each fixed $i$, we have $|x_i^{(n)}-x_i^{(m)}|<\sqrt{\epsilon}$. Since $\mathbb{R}$ is complete we each $x_i^{(n)}\to x_i$ as $n\to\infty$. Because of this, we can substitute $x_i$ for $x^{(m)}_i$ in our inequality (1), giving us: \begin{equation}||X^n-X||=\sum_{i=1}^{\infty}(x_i^{(n)}-x_i)^2\leq\epsilon \tag{2}\end{equation} This means that $(X^n-X)\in\ell^2$. Since $X=X^n+(X-X^n)$ then $X\in\ell^2$. Then by equation (2), $X^n\to X$.
 A: There are many problems. For instance, asserting that $\|X^n-X^m\|<\varepsilon$ does not mean that, for each fixed $i$, $(x_i^{\,n}-x_i^{\,m})^2<\varepsilon$. The fact that all norms on $\Bbb R$ are equivalent is not relevant here, since you are not working on $\Bbb R$. And you mention certain numbers $\varepsilon_i$ without explainig what they are.
You will find here a proof that $\ell^p$ is complete.
A: There is a problem just after "We can take this sequence to be such that $\epsilon > \sum \epsilon_i$". Presumably you mean we are given $\epsilon$ and we take the sequence $\epsilon_1, \epsilon_2, \ldots$ so the inequality holds. So good so far. Then claim we can get all $(x^n_i - x_i) \le \epsilon_i$ simultaneously but do not prove it. This is the problem.
It is certainly true that $$(x^n_1 - x_1) \le \epsilon_1\qquad  (x^n_2 - x_2) \le \epsilon_2 \qquad  \ldots \qquad   (x^n_m - x_m) \le \epsilon_n  $$ for all $n \ge \max\{N_1,N_2,\ldots, N_m\}$. But this only gives finitely many coordinates.
Proof Hint: You have not used how $X^n$ are in $\ell^2$. In particular how $\sum_i (x^n_i) ^2 < \infty$.
