$\mathbb{R}^n$ is a complete metric space.

Consider a Cauchy sequence $\{\mathbf{x}_k\}$ in $\mathbb{R}^n$, we want to show it converges to a point $\mathbf{x} \in \mathbb{R}^n$. That is to say, if $|\mathbf{x - x_k}| \to 0$ as $k \to \infty$.

Hence, we let $\epsilon \to 0$, and we get $\mathbf{|x_k - x_j|} < \epsilon$ by Cauchy sequence, and let $\mathbf{x = x_j}$ we showed the desired result.

Definition $\mathbb{R}^n$ is a complete metric space. Every Cauchy sequence in $\mathbb{R}^n$ converges to a point of $\mathbb{R}^n$.

Definition Cauchy sequence. Given $\epsilon > 0$, there is an integer $K$ such that $\mathbf{|x_k - x_j|} < \epsilon$ for all $k,j \geq K$.

I am not fond of my proof, because I am not certain if I can approach $\epsilon$ to be zero, nor if I can equate $\mathbf{x}$ to be $\mathbf{x_j}$ since $\mathbf{x_j}$ is changing while $\epsilon$ changes.

Edit Especially, I am baffled that why we need to do it in coordinates? I think they can be subtracted directly, as the definition of Cauchy sequence I added a short while ago?


Every Cauchy sequence is bounded, hence contained in a compact cube, admits thus a convergent subsequence and therefore converges itself.

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    $\begingroup$ Indeed, that's it. The first and last statement are quite easy to prove from the definition, the second follows from the Heine-Borel theorem (not so easy), the third is a standard criterion for compactness $\endgroup$ – user88576 Aug 21 '13 at 21:35
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    $\begingroup$ Technically, you only need the Bolzano-Weierstrass theorem. That being said, either proof relies on the monotone convergence theorem, which is a slightly simpler version of this problem. $\endgroup$ – Omnomnomnom Aug 21 '13 at 21:48
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    $\begingroup$ Also, you implicitly assume that a Cauchy sequence with a convergent subsequence converges, which doesn't really work given the framework of the problem $\endgroup$ – Omnomnomnom Aug 21 '13 at 21:50
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    $\begingroup$ Your proof depends on the completeness of $\mathbb R$ because the B-W and Heine Borel proof depend on the completeness of $\mathbb R$, because the monotone convergence theorem depends on the completeness of $\mathbb R$. That being said, it is interesting that with your framework, you can reduce the original problem to that of a Cauchy sequence with a convergent subsequence, and that is an admirable insight. I originally thought that this proof was somehow circular, but after reviewing the finer points, I guess I was wrong. $\endgroup$ – Omnomnomnom Aug 21 '13 at 23:12
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    $\begingroup$ @Sum Ting Wong: the fact that every (nonempty) bounded subset of $\mathbb{R}$ has a supremum is precisely the completeness of $\mathbb{R}$. $\endgroup$ – Pete L. Clark Sep 4 '13 at 22:14

This proof isn't quite right.

How did we prove that Cauchy sequences converge in $\mathbb R$? In fact, we didn't, this is just supposed to follow from the construction of the real numbers. Try to see the shortcomings of your method in this context. Or, note that any proof of completeness should fail on $(0,1)$ with the sequence $x_k=\frac1k$, and see why your method doesn't raise any red flags where it should.

How can we use the completeness of $\mathbb R$ to deduce the completeness of $\mathbb R^n$?

Hint: try considering the sequence that you get by looking at a particular coordinate of $\mathbf x_k$. Why is this sequence Cauchy, and how does that help?

Answer: given $\mathbf x_j\in\mathbb R^n$, let $x_{i,j}$ be the $i^{th}$ coordinate of $\mathbf x_j$. For each $i$, $\{x_{i,j}\}_{j\geq1}$ is a Cauchy sequence in $\mathbb R$ (Because if one component does not converge, the norm does not.), and thus converges in $\mathbb R$. Let $x_i$ be the limit of this sequence.

Now, let $\mathbf x=(x_1,x_2,\dots,x_n)$ with $x_i$ as defined above.

Consider any $\epsilon>0$. By the convergence of the coordinate sequences, we may select an integer $K$ so that for each $j>K:|x_{i,j}-x_i|<\epsilon/n$. We note that for $j>K:$ $$ \|\mathbf x_j-\mathbf x\|=\\ \|((x_{1,j}-x_1),(x_{2,j}-x_2),\dots,(x_{n,j}-x_n))\| \leq\\ |x_{1,j}-x_1|+|x_{2,j}-x_2|+\dots+|x_{n,j}-x_n)| <\\ \epsilon/n + \epsilon/n + \dots + \epsilon/n = \epsilon $$ Thus, $\mathbf x_j\to\mathbf x$, which means that an arbitrary Cauchy sequence must converge.

  • $\begingroup$ Oh, after looking at the definition for a second time, I think we are defined on $\mathbb{R}^n$, hence we won't need to consider coordinate-wisely? $\endgroup$ – 1LiterTears Aug 21 '13 at 20:55
  • $\begingroup$ I've updated my answer to explain what I meant. Hope that's clearer $\endgroup$ – Omnomnomnom Aug 21 '13 at 21:51
  • $\begingroup$ $\|$ means norm, $|$ means absolute value, right? $\endgroup$ – 1LiterTears Aug 21 '13 at 22:08
  • $\begingroup$ Yes, that's right! Also, the edit was right. $\endgroup$ – Omnomnomnom Aug 21 '13 at 22:59
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    $\begingroup$ We need to do it in coordinates because given the Cauchy sequence, we need to find the coordinates of its limit. $\endgroup$ – Omnomnomnom Aug 21 '13 at 23:14

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