# Completion of a metric space

Let $R$ be a topological ring (i.e addition and product are continuous) which we assume it is metrizable with metric d and consider the completion $\hat{R}$ of the ring $R$ defined as the set of all classes of equivalent Cauchy sequences. Question: why is the completion of $\hat{R}$ (defined in this way) also a topological ring?

It is certainly metrizable by the standard metric: $\hat{R}$, $\hat{d}([(x_{n})],[(y_{n})])=\operatorname{lim} d(x_{n},y_{n})$.

To check addition is continuous: suppose $([(x_{n})],[(y_{n})])$ is a sequence of elements in $\hat{R} \times \hat{R}$ such that$([(x_{n})],[(y_{n})]) \rightarrow ([x],[y])$ then by by definition of convergence in a metric product space we have that:

$[(x_{n})] \rightarrow [x]$ and $[(y_{n})] \rightarrow y$

Therefore $d(x_n,x) \rightarrow 0$ and $d(y_n,y) \rightarrow 0$ as $n \rightarrow \infty$. Since we are in metric spaces this says $x_{n} \rightarrow x$ and $y_{n} \rightarrow y$ as $n \rightarrow \infty$. By assumption $R \times R$ is continuous with respect addition so $x_{n} + y_{n} \rightarrow x+y$.

But then $\hat{d}([(x_{n}+y_{n})])=[x+y]$ so we have that addition is continuous.

Is this OK?

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This is false if you don't posit some additional relationship between the metric and the structure maps of the ring beyond that they are continuous. Consider $\mathbb{R}$ equipped with the metric $$d(x, y) = |\arctan x - \arctan y|$$

(where $\arctan x \in (-\frac{\pi}{2}, \frac{\pi}{2})$). This induces the same topology as the usual metric so addition and multiplication are still continuous, but the completion with respect to this metric is the extended real line $\mathbb{R} \cup \{ +\infty, -\infty \}$ and it is not possible to extend either the addition or the multiplication.

By assumption $R \times R$ is continuous with respect addition so $x_n + y_n \to x + y$.

$x + y$ is not well-defined at this step in the argument if either $x$ or $y$ does not lie in $R$. What you probably meant is something like this: the sum of Cauchy sequences is Cauchy, so we can define $x + y$ by taking $\lim (x_n + y_n)$.

Unfortunately, the continuity of addition does not guarantee that the sum of Cauchy sequences is Cauchy; in the above example, take $$x_n = (-1)^n + n, y_n = (-1)^n - n.$$

Even when the sum of two Cauchy sequences is Cauchy, it's false that $\lim (x_n + y_n)$ is uniquely determined by $\lim x_n$ and $\lim y_n$; to see this, take $$x_n = -n, y_n = n, x_n = -n, y_n = n^2.$$

Edit: for the $I$-adic completion things are considerably simpler because we can write $d(x, y) = |x - y|$ for some function $| \cdot | : A \to \mathbb{R}$ satisfying (among other things) $|x + y| \le |x| + |y|$ and $|xy| \le |x| |y|$. These properties readily imply all of the nice properties that you want.

In my opinion the conceptually cleanest way to think about $I$-adic completion is not using Cauchy sequences but as follows: the $I$-adic completion is the inverse limit of the quotients $R/I^n$.

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Perhaps one should assume that addition and multiplication are uniformly continuous. –  Nate Eldredge Apr 10 '12 at 2:32
@Nate: yes, this seems like a good idea. –  Qiaochu Yuan Apr 10 '12 at 2:34
@Mertier: with those extra properties all of the arguments that didn't work before work now. Just try it! It's a nice exercise to show that sums and products of Cauchy sequences are Cauchy with these extra properties and then you can try to prove that sum and product are well-defined on the completion. –  Qiaochu Yuan Apr 10 '12 at 3:29
@Qiaochu Yuan: I get confused. Why showing sum and product are well defined implies continuity? please don't think I want you to solve my problem I just want to understand what exactly I need to prove and this is what confuses me. –  Mertier R. Apr 10 '12 at 6:56
@Mertier: well, I assume you want to prove that the $I$-adic completion is a topological ring? So show that the sum and product of Cauchy sequences is Cauchy, show that the limits of a sum and product depend only on the limits of the summands and productands, show that they are continuous in the $I$-adic topology. Right? –  Qiaochu Yuan Apr 10 '12 at 16:21