# Show $\mathbb R^n$ is complete.

Show $\mathbb R^n$ is complete.

At this point, I am trying to work through the problem in my textbook, there is one step that I do not understand and would like explained. Here's my proof so far:

Let $x_k = \{\xi_1^{(k)},\xi_2^{(k)},\ldots,\xi_n^{(k)}\}$ be a Cauchy sequence in $\mathbb R^n$. Then by definition, given $\varepsilon > 0$ there exists $N$ such that if $s, t > N$ then \begin{align*} d(x_s, x_t)&< \varepsilon\\ \Bigg[\sum_{j=1}^n \bigg(\xi_j^{(s)} - \xi_j^{(t)}\bigg)\Bigg]^{\frac12}&< \varepsilon\\ \sum_{j=1}^n \bigg(\xi_j^{(s)} - \xi_j^{(t)}\bigg)^2&< \varepsilon^2 \end{align*} which implies that $\bigg(\xi_j^{(s)} - \xi_j^{(t)}\bigg)^2 < \varepsilon^2$ or equivalently $\bigg\vert \xi_j^{(s)} - \xi_j^{(t)} \bigg\vert < \varepsilon$ for $j = 1, 2, \ldots, n$. Hence $\xi_j^{(k)}$ is a Cauchy sequence of real numbers. Moreover since $\mathbb R$ is complete, every Cauchy sequence of real numbers is convergent. So it follows that $\xi_j^{(k)} \to \xi_j$. That is $\lim_{k \to \infty} \xi_j^{(k)} = \xi_j \in \mathbb R$ for $j = 1, 2, \ldots, n$. Set $x = (\xi_1, \xi_2, \ldots, \xi_n)$ and observe that $x \in \mathbb R^n$.

Here is where I get stuck; the book then does

Since $d(x_s, x_t) < \varepsilon$, letting $t \to \infty$ we get $d(x_s, x) < \varepsilon$. Thus by definition $x_k \to x \in \mathbb R^n$ and we can conclude that $\mathbb R^n$ is complete.

My question is how we can just let $t \to \infty$ and get $x$? Using that reasoning couldn't we just say letting $k \to \infty$ we get $x$?

Moreover, a bit of a subquestion, my definition of Cauchy sequence is:

A sequence $(x_n)$ in a metric space $X = (X, d)$ is said to be Cauchy if for every $\varepsilon > 0$ there is an $N = N(\varepsilon)$ such that $d(x_m, x_n) < \varepsilon$ for every $m, n > N$.

Is this the same as:?

A sequence $(x_n)$ in a metric space $X = (X, d)$ is said to be Cauchy if for every $\varepsilon > 0$ there is an $N = N(\varepsilon)$ such that $d(x_s, x_t) < \varepsilon$ for every $s, t > N$.

Also, I am using Kreyszig's Introductory Functional Analysis and this is example on page 33.

-
Those last two definitions are clearly equivalent; you just used $m,n$ in one of them and $s,t$ in the other. – Benjamin Dickman Feb 3 '13 at 10:11
Thanks, I was getting confused with the naming of the sequence $(x_n)$ and using that $n$ in the definition. – Robert Cardona Feb 3 '13 at 10:13
Ah yes, strictly speaking the latter definition is better. Or change the $(x_n)$ to $(x_k)$ in the first one. Otherwise you are right: it is confusing and somewhat incorrect. – Benjamin Dickman Feb 3 '13 at 10:15
Ok, thanks! I think the two main books I'm working through use the first notation, I just wanted to make sure the second approach was still correct and that there wasn't some hidden meaning in using $n$ twice that I couldn't see. – Robert Cardona Feb 3 '13 at 10:17

For any $\varepsilon$, select $N_i$ such that if $s>N_i$ then $|\xi^s_i - \xi_i| < \frac{\varepsilon}{\sqrt{n}}$. This can be done since $\xi_i^s\to\xi_i$. Let $N = \max\{N_1,\ldots,N_n\}$. Therefore if $s>N$, then $$d(x_s,x)=\sqrt{\sum_1^n(\xi^s_i-\xi_i)^2}<\varepsilon$$
Well they explicitly say "since $\mathbb{R}$ is complete," so presumably they first explained that $\mathbb{R}$ is complete elsewhere and are now using that fact to show that $\mathbb{R}^n$ is also complete. – user38355 Feb 25 '13 at 12:51
At the stage they say, $t \to \infty$ and replace $x_t$ with $x$, they're assuming $\{x_t\}$ converges to $x$. Circular reasoning. – genepeer Feb 25 '13 at 12:54
"$\mathbb{R}$ is complete" was only used to show that the sequence of components $\xi_i^s \to \xi_i$. – genepeer Feb 25 '13 at 12:57
No, they define $x$ to be the vector whose coordinates are the limits of the coordinates of $x_k$ which exist since they showed the coordinates to be Cauchy sequences and $\mathbb{R}$ is complete. Then they write down the inequality from before. Then they take the limit of the individual coordinates as $t \to \infty$ in the inequality to get a new inequality. Then they see that this inequality actually reads $d(x_s,x) < \epsilon$. Now they conclude that actually $x_k \to x$ in the metric of $\mathbb{R}^n$. No circular reasoning. – user38355 Feb 25 '13 at 20:24
So why even bother with the inequality since clearly $x_k$ will tend to $x$ for just the same reason $x_t$ does. Why didn't they say, "Take the limit of the individual coordinates of ${x_k}$ (original sequence) as $k\to \infty$ and you get $x$." – genepeer Feb 25 '13 at 22:03