Theorem 1

Let $(\mathbf{x}_n)_{n \in \mathbf{N}}$ be a sequence of points of a complete normed algebra $A$.

a) If $(\mathbf{x}_n)$ is multipliable and has for product an invertible element of $A$, for every $\varepsilon > 0$, there exists a finite subset $J_0$ of $\mathbf{N}$ such that, for every finite subset $L$ of $\mathbf{N}$ not meeting $J_0$, one has $\| \mathbf{e} - \mathbf{p}_L \| \leq \varepsilon$.

b) Conversely, if the sequence $(\mathbf{x}_n)$ satisfies this condition, it is multipliable. Moreover, if each of the $\mathbf{x}_n$ is invertible, $\prod_{n \in \mathbf{N}} \mathbf{x}_n$ is invertible.

a) Let $\mathbf{p}$ be the product of the multipliable sequence $(\mathbf{x}_n)$, and suppose that $\mathbf{p}$ is invertible in $A$; then (IX, p. 40, prop. 14), there exist $\alpha > 0$ and $a > 0$ such that, for every $y \in A$ such that $\| y - \mathbf{p} \| \leq \alpha$, $y$ is invertible, and $\| y^{-1} \| \leq a$. By assumption, for every $\varepsilon$ such that $0 < \varepsilon < \alpha$, there exists a finite subset $H_0$ of $\mathbf{N}$ such that, for every finite subset $H$ of $\mathbf{N}$ containing $H_0$, one has $\| \mathbf{p}_H - \mathbf{p} \| \leq \varepsilon$. Let $J_0 = \{0, m\}$ be an interval of $\mathbf{N}$ containing $H_0$; for every finite subset $L$ of $\mathbf{N}$ not meeting $J_0$ the integers belonging to $L$ are all greater than those belonging to $H_0$; therefore, if one sets $H = H_0 \cup L$, one has $\mathbf{p}_H = \mathbf{p}_{H_0} \mathbf{p}_L$. Now, since $\| \mathbf{p}_{H_0} - \mathbf{p} \| \leq \varepsilon \leq \alpha$, $\mathbf{p}_{H_0}$ is invertible, and one has $\| \mathbf{e} - \mathbf{p}_{H_0}^{-1} \mathbf{p} \| \leq \varepsilon \| \mathbf{p}_{H_0}^{-1} \| \leq a \varepsilon$; from the relation $\| \mathbf{p}_{H_0} \mathbf{p}_L - \mathbf{p} \| \leq \varepsilon$, one deduces $\| \mathbf{p}_L - \mathbf{p}_{H_0}^{-1} \mathbf{p} \| \leq \varepsilon \| \mathbf{p}_{H_0}^{-1} \| \leq a \varepsilon$ and finally $\| \mathbf{e} - \mathbf{p}_L \| \leq 2a \varepsilon$.

b) Suppose that, for every $\varepsilon > 0$, there exists a finite subset $J_0$ of $\mathbf{N}$ such that, for every finite subset $L$ of $\mathbf{N}$ not meeting $J_0$, one has $\| \mathbf{e} - \mathbf{p}_L \| \leq \varepsilon$. Let $H_0 = \{0, p\}$ be an interval of $\mathbf{N}$ containing $J_0$; every finite subset $H$ of $\mathbf{N}$ containing $H_0$ can be written $H_0 \cup L$, where the integers belonging to $L$ are greater than those belonging to $H_0$; we therefore have $\mathbf{p}_H = \mathbf{p}_{H_0} \mathbf{p}_L$, and since $L$ does not meet $J_0$, $\| \mathbf{p}_H - \mathbf{p}_{H_0} \| \leq \varepsilon \| \mathbf{p}_{H_0} \|$, and consequently $\| \mathbf{p}_H \| \leq (1 + \varepsilon) \| \mathbf{p}_{H_0} \|$. If $\mathbf{p}_{H_0} = 0$, the sequence $(\mathbf{x}_n)$ is obviously multipliable and has product 0; excluding this trivial case, there exists an interval $H_1 = \{0, q\}$, containing $H_0$ and such that, for every finite subset $L$ of $\mathbf{N}$ not meeting $H_1$, one has $\| \mathbf{e} - \mathbf{p}_L \| \leq \varepsilon (\| \mathbf{p}_{H_0} \|)^{-1}$. We deduce as above that, for every finite subset $H \supset H_1$

The Cauchy criterion therefore shows that $J \mapsto \mathbf{p}_J$ has a limit in $A$ following the filtered set $\mathscr{F}(\mathbf{N})$.

If all the $x_n$ are invertible, the same is true of all the finite partial products $p_J$; for every finite subset $H$ containing $H_0$, one can therefore write

this shows that, in the multiplicative group $G$ of the invertible elements of $A$, the image by the mapping $J \mapsto p_J$ of the filter of sections of $\mathfrak{F}(N)$ is a Cauchy filter basis for the left uniform structure of the group $G$; but since $G$ is complete ($IX$, p. 40, prop. 14), the limit of the mapping $J \mapsto p_J$ belongs to $G$.