I'm trying to prove $l_p$ spaces are complete.
We have an $l_p$ space $W$. Let us take a cauchy sequence. There exists $N_0\in\Bbb{N}$ such that for $m,n>N_0$, $d(x^m,x^n)<\epsilon$. This implies $\left (|x_1^m-x_1^n|^p + |x_2^m-x_2^n|^p+\dots\right)^{\frac{1}{p}}<\epsilon$. This in turn implies $|x^m_j-x^n_j|<\epsilon$ for any $j\in\Bbb{N}$.
Hence, if any $p\in W$, is part of a cauchy sequence in $W$, every coordinate of $P$ is also in a cauchy sequence in $\Bbb{R}$. As $\Bbb{R}$ is complete, every such cauchy sequence has a limit in $\Bbb{R}$. Let the limit of the $i^{th}$ coordinate be $l_i$. ALso, let $L=\{l_1,l_2,l_3,\dots\}$.
Let us come back to $d(x^m,x^n)$. Keeping $n$ constant, keep on increasing $m$. For every $m+i$, where $i\in\Bbb{N}$, $d(x^{m+i},x^n)<\epsilon$.
My book says
Hence, $d(L,x_n)<\epsilon$
I don't see how that follows from the fact that $d(x^{m+i},x^n)<\epsilon$. Howveer high the value of $i$ is, $x^{m+i}$ can never equal $L$.
My thoughts were $d(L,x^n)\leq d(L,x^{m+i})+d(x^{m+i},x^n)$. However, I don't know how to make $d(L,x^{m+i})$ arbitrarily small. The reason for this is $d(L,x^{m+i})=\left(|l_1-x_1^{m+i}|^p+|l_2-x_2^{m+i}|^p+\dots\right)^{\frac{1}{p}}$. Although each $l_j-x_j^{m+i}$ can be made arbitrarily small, there are infinite coordinates. Hence, if $|l_j-x_j^{m+i}|<\epsilon$ for $m+i>N_j$, one can't find $\max\{N_1,N_2,\dots\}$. ALso, even if $|l_j-x_j^{m+i}|$ can be made arbitrarily small, there are an infinite number of these numbers to add.
- Is the argument in the book justified? If it is, how?
- How can I proceed with my argument?
Thanks in advance!