Prove $\mathbb{R}^n$ space is complete - Summation Notation

Question

I’m studying the following prove that $\mathbb{R}^n$ is complete from Kreyzig’s Introduction to Functional Analysis.

Wonder if I could get some help to clarify a few things to confirm that I understand it correctly.

From (1) below in the proof. I wanted to make sure I understand the expression and read the notation correctly.

a) Are both ξ_(j )^((m) ) and ξ_(j )^((r) ) are from the same Cauchy sequence?
b) For example, let’s say using (1/n) as a Cauchy sequence. Let N = 100. With m = 101, n = 200. Does that mean for the summation index j = 1, ξ_(j )^((m) )=  ξ_102 and ξ_(j )^((r) )=  ξ_201? And where if j = n, ξ_(j )^((m) )=  ξ_(101+n) and ξ_(j )^((r) )=  ξ_(200+n)?

Prove that Euclidean space $\mathbb{R}^n$ is complete. Metric on $\mathbb{R}^n$ is defined by $$d(x,y)=\left(\sum_{j=1}^n (\xi_j-\eta_j)^2\right)^{1/2}$$ where $x=(\xi_j)$ and $y=(\eta_j)$.

We consider any Cauchy sequence $(x_m)$ in $\mathbb{R}^n$, writing $$x_m=\left(\xi_{1}^{(m)}\cdots,\xi_{n}^{(m)}\right).$$

Since $(x_m)$ is Cauchy, for every $ε>0$ there is an $N\in\mathbb{N}$ such that $$(1)\qquad d(x_m,x_r)=\left(\sum_{j=1}^n \left(\xi_{j}^{(m)}-\xi_{j}^{(r)}\right)^2\right)^{1/2}<\epsilon\qquad(m,r>N)$$

Thank you!

Answer 1

It seems to me that you're confused because your Cauchy sequence $(x_m)$ is a sequence of "sequences" (in $\mathbb{R}^n)$. The symbol $\xi_j^m$ denotes the $j$-th component of the point $x_m$ in $R^n$, that is, $x_m:=(\xi_1^m,\xi_2^m,\ldots,\xi_n^m)$. The upper index gives the $m$-th term of the sequence $(x_m)$ and that $m$-th term happens to have $n$ components $\xi_1^m,\xi_2^m,\ldots,\xi_n^m$. Perhaps writing $(x^m)$ for your sequence would be more consistent with the notation chosen, so let's take that convention from now on.

It might help to think of the sequence $(x^s)$ (let's take a different index $s$ not to confuse things) as an infinite matrix where the $m$-th row represents its $m$-th term: $$ \begin{pmatrix}x^1\\x^2\\x^3\\\vdots\end{pmatrix}=\begin{pmatrix}\xi_1^1&\xi_2^1&\xi_3^1&\cdots&\xi_n^1\\\xi_1^2&\xi_2^2&\xi_3^2&\cdots&\xi_n^2\\\xi_1^3&\xi_2^3&\xi_3^3&\cdots&\xi_n^3\\\vdots&\vdots&\vdots&\ddots&\vdots\end{pmatrix} $$

a) Are both $\xi_j^m$ and $\xi_j^r$ from the same Cauchy sequence?

Well, $\xi_j^m$ is the $j$-th component of $x^m$ and $ξ_j^r$ is the $j$-th component of $x^r$. $x^m$ and $x^r$ are from the same Cauchy sequence $(x^s)$.

b) For example, let’s say we are using $(1/n)$ as a Cauchy sequence. Let $N = 100$. With $m = 101,n = 200$. Does that mean for the summation index $j = 1$, $\xi_j^m=\xi_{102}$ and $\xi_j^r=\xi_{201}$? And what if $j=n,\xi_j^m=\xi_{101+n}$ and $\xi_j^r=\xi_{200+n}$?

I don't really understand this question. $N$ doesn't appear in the summation. We have $$ \left(\sum_{j=1}^n(\xi_j^m-\xi_j^r)^2\right)^{1/2}=\left((\xi_1^m-\xi_1^r)^2+(\xi_2^m-\xi_2^r)^2+\cdots+(\xi_n^m-\xi_n^r)^2\right)^{1/2} $$ (I suppose you forgot to square $\xi_j^m-\xi_j^r$ and that $d$ is the metric issued from the euclidean norm $\|\cdot\|_2$ in $\mathbb{R}^n$).

If for a given $\epsilon$ the integer $N=100$ works and if $m=101$ and $r=200$, then the Cauchy condition says, in particular, since $m,r>N$ (you typed $m,n>N$ in your question but that's a typo, it should be $r$ instead of the fixed dimension $n$ of $\mathbb{R}^n$), that the distance $d(x^m,x^r)$ between the $m$-th and the $r$-th terms of the Cauchy sequence $(x^s)$ is less than $\epsilon$. That is, \begin{align} d(x^{101},x^{200})&=\left(\sum_{j=1}^n(\xi_j^{101}-\xi_j^{200})^2\right)^{1/2}\\&=\left((\xi_1^{101}-\xi_1^{200})^2+(\xi_2^{101}-\xi_2^{200})^2+\cdots+(\xi_n^{101}-\xi_n^{200})^2\right)^{1/2}\\&<\epsilon\quad\quad\quad\quad\text{since }101,200>100 \end{align}