Proposition 13

Let G be the group of units of a complete normed algebra A. Then G is an open subset of A; the topology induced on G by the topology of A is compatible with the group structure of G; and G, endowed with this topology, is a complete group (with respect to each of its two uniformities).

The corollary to Proposition 12 shows that G contains a neighbourhood V of $e$ in A; hence, for each $x_0 \in G$, the elements of $x_0V$ are units, and $x_0V$ is a neighbourhood of $x_0$ in A, since $x \mapsto x_0x$ is a homeomorphism of A onto itself ($x_0$ being a unit of A). Hence G is open in A.

To show that the topology induced on G by the topology of A is compatible with the group structure of G, it is sufficient to show that the function $x^{-1}$ is continuous on G. Let $x_0 \in G$, and for each $x \in G$, write $x$ in the form $x = x_0(e + u)$, so that $u = x_0^{-1}(x - x_0)$; then $||u|| \leq ||x_0^{-1}|| \cdot ||x - x_0||$, and thus if $||x - x_0|| \leq 1 / ||x_0^{-1}||$, we have $||u|| \leq 1$, $e + u = x_0^{-1}x$ is a unit, the series whose general term is $(-1)^n u^n$ is absolutely convergent, and

from which it follows that

As $x$ tends to $x_0$, $||x - x_0||$ tends to 0, and since

remains bounded, it follows that $x^{-1}$ tends to $x_0^{-1}$.

Finally, to show that the left uniformity of G is complete, let us show that every Cauchy filter $\tilde{\mathscr{F}}$ with respect to this uniformity is a Cauchy filter with respect to the additive uniformity of $A$ and converges to a point of $G$. For each $\varepsilon$ such that $0 < \varepsilon < 1$, there is a set $M \in \mathfrak{F}$ such that $||x^{-1}y - e|| \leq \varepsilon$ for all $x, y$ in $M$, i.e., such that

Let $a$ be a point of $M$; for each $x \in M$ we have $||x - a|| \leq \varepsilon \ ||a||$, and therefore $||x|| \leq (1 + \varepsilon)||a||$. On the other hand, there exists a set $N \subset M$ belonging to $\mathfrak{F}$ and such that $||x^{-1}y - e|| \leq \varepsilon/(1 + \varepsilon)||a||$ for all $x$ and $y$ in $N$; it follows that $||y - x|| \leq \varepsilon ||x||/(1 + \varepsilon)||a|| \leq \varepsilon$, which shows that $\mathfrak{F}$ is a Cauchy filter with respect to the additive uniformity of $A$, and therefore converges to a point $x_0$, since $A$ is complete. Since $x_0$ is the limit of $\mathfrak{F}$, we have $||x^{-1}x_0 - e|| \leq \varepsilon$ for all $x \in M$ by the principle of extension of inequalities; since $\varepsilon > 1$, it follows that $x^{-1}x_0$ is a unit in $A$; hence $x_0$ is a unit, i.e., $x_0 \in G$.