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?