Proving $\ell^p$ is complete

Solution 1:

Let $\left( x^{(n)}\right)_{n=1}^{\infty} \subset \ell^p$ be a Cauchy sequence. Since I see you have troubles with your notations of sequence of sequences, this is the notation that I will use for each element $x^{(n)}$ in the sequence: $$ x^{(n)} = \left( x_j^{(n)}\right)_{j=1}^{\infty} = \left( x_1^{(n)},x_2^{(n)}, \cdots \right)\in \ell^p $$

For $x= \left( x_j\right)_{j=1}^{\infty} , y= \left( y_j\right)_{j=1}^{\infty} \in \ell^p$, lets define the $p$-norm $\| \cdot \|_p$ as the one who induces $d_p$, that is $\|x-y\|_p=d_p(x,y)$. Precisely $$ \|x-y\|_p= \left(\sum_{j=1}^{\infty} \left|x_j-y_j\right|^p\right)^{1/p} $$

Now lets continue, take $\varepsilon>0$, then there exist a $N=N(\varepsilon) \in \mathbb{N}$, such that if $m,n >N$ then $$ \|x^{(m)}-x^{(n)}\|_p<\varepsilon. $$ Thus, for any $j \in \mathbb{N}$, it follows that $$ \left|x^{(m)}_j-x^{(n)}_j\right|^p \leq \sum_{j=1}^{\infty} \left|x^{(m)}_j-x^{(n)}_j\right|^p = \|x^{(m)}-x^{(n)}\|^p_p<\varepsilon^p $$ that is, for any $j \in \mathbb{N}$ the sequence $\left( x^{(n)}_j\right)_{n=1}^{\infty} \subset \mathbb{R}$ is a Cauchy one. Since $\mathbb{R}$ is complete, for each $j$ there exist a $x_j \in \mathbb{R}$ such that $$ \lim_{n \to \infty} x^{(n)}_j = x_j $$ Lets fix $k \in \mathbb{N}$, then in a similar way for $m,n >N$ \begin{equation} \sum_{j=1}^{k} \left|x^{(m)}_j-x^{(n)}_j\right|^p \leq \sum_{j=1}^{\infty} \left|x^{(m)}_j-x^{(n)}_j\right|^p = \|x^{(m)}-x^{(n)}\|^p_p<\varepsilon^p \tag{1} \end{equation} Letting $n \to \infty$ in (1), we get that for $m>N$ \begin{equation} \sum_{j=1}^{k}\left|x^{(m)}_j-x_j\right|^p < \varepsilon^p \tag{2} \end{equation} Then by the usual triangle inecuality ( Minkowski's inequality for $\|\cdot\|_p$ in $\mathbb{R}^k$) we get that if $m>N$ $$ \left( \sum_{j=1}^{k}|x_j|^p \right)^{1/p} \leq \left( \sum_{j=1}^{k}\left|x^{(m)}_j-x_j\right|^p \right)^{1/p} + \left( \sum_{j=1}^{k} \left|x^{(m)}_j \right| \right)^{1/p} < \varepsilon + \left( \sum_{j=1}^{k} \left|x^{(m)}_j \right| \right)^{1/p} $$ by letting $k \to \infty$, we get $\|x\|_p\leq \varepsilon + \|x^{(m)}\|_p$, which is the same as getting that $x=\left( x_j\right)_{j=1}^{\infty} \in \ell^p$. Again, letting $k \to \infty$ in (2), we obtain that if $m>N$ $$ \|x^{(m)}-x\|_p^p= \sum_{j=1}^{\infty}\left|x^{(m)}_j-x_j\right|^p < \varepsilon^p $$ thus $$ \lim_{m \to \infty} \|x^{(m)}-x\|_p= 0 $$ so indeed, $\left( x^{(m)}\right)_{m=1}^{\infty} \subset \ell^p$, is a convergent sequence who converges to $x \in \ell^p$. We conclude then that $\ell^p$ is a complete metric space for $1\leq p < \infty$.