Extending multiplication to the completion of an algebra

Question

Let $A$ be an algebra with a submultiplicative norm $\|\cdot\|$, that is, assume $$ \|ab\| \leq \|a\|\|b\|, ~~~~~~ \text{ for all } a,b \in A. $$ Let us denote by $\overline{A}$ the completion of $A$ with respect to $\|\cdot\|$. I have read that the multiplication extends to a multiplication on the completions, but am having trouble writing down a precise proof.

Answer 1

For $a, b \in \overline{A}$, let these be the limits of Cauchy sequences $a_n$ and $b_n$ in $A$. Then you can define $ab = \lim_{n \to \infty} a_n b_n$ . To show that the limits do not depend on the choice of the Cauchy sequence, suppose $a'_n$ and $b'_n$ are other Cauchy sequences with limits $a$ and $b$ respectively, and note that

$$ \|a_n b_n - a'_n b'_n\| = \|a_n b_n - a'_n b_n + a'_n b_n - a'_n b'_n\| \le \|a_n - a'_n\| \|b_n\| + \|a'_n\| \|b_n - b'_n\|$$

and $\|a_n - a'_n\| \to 0$, $\|b_n - b'_n\| \to 0$, $\|a'_n\| \to \|a\|$ and $\|b_n\| \to \|b\|$.