Theorem 1

Let (x_n)_{n \in \mathbf{N}} be a sequence of points in a complete normed algebra A. a) If (x_n) is multipliable and if its product is a unit of A, then for each $\varepsilon > 0$ there exists a finite subset J_0 of \mathbf{N} such that, for every finite subset L of \mathbf{N} which does not meet J_0, we have $\|e - p_L\| \leq \varepsilon$. b) Conversely, if the sequence (x_n) satisfies this condition, it is multipliable. Moreover, if each x_n is a unit, then $\prod_{n \in \mathbf{N}} x_n$ is a unit.

a) Let p be the product of the multipliable sequence (x_n), and suppose that p is a unit in A; then (Chapter IX, § 3, no. 7, Proposition 13) there exist $\alpha > 0$ and $a > 0$ such that, for all $y \in A$ for which we have

y is a unit and $\|y^{-1}\| \leq a$. By hypothesis, for every $\varepsilon$ such that $0 < \varepsilon < \alpha$, there is a finite subset H_0 of \mathbf{N} such that, for every finite subset H of \mathbf{N} containing H_0, we have $\|p_H - p\| \leq \varepsilon$. Let $J_0 = [0, m]$ be an interval of \mathbf{N} which contains H_0; for each finite subset L of \mathbf{N} which does not meet J_0, the integers belonging to L are all greater than those belonging to H_0; hence, if $H = H_0 \cup L$, we have $p_H = p_{H_0} p_L$. Now, since $\|p_{H_0} - p\| \leq \varepsilon \leq \alpha$, $p_{H_0}$ is a unit, and since $\|p_{H_0} p_L - p\| \leq \varepsilon$, we deduce and finally $\|e - p_L\| \leq 2a \varepsilon$.

b) Suppose that, for each $\varepsilon > 0$, there exists a finite subset J_0 of \mathbf{N} such that, for every finite subset L of \mathbf{N} which does not meet J_0, we have $\|e - p_L\| \leq \varepsilon$. Let $H_0 = [0, p]$ be an interval of \mathbf{N} which contains J_0; then every finite subset H of \mathbf{N} which contains H_0 can be written in the form $H_0 \cup L$, where the integers in L are all greater than those in H_0; hence we have $p_H = p_{H_0} p_L$, and since L does not meet $J_0$, $\|p_H - p_{H_0}\| \leq \varepsilon \|p_{H_0}\|$, and consequently $\|p_H\| = (1 + \varepsilon)\|p_{H_0}\|$. If $p_{H_0} = 0$, the sequence $(x_n)$ is evidently multipliable and its product is 0; excluding this trivial case, there is an interval $H_1 = [0, q]$ containing $H_0$ and such that, for every finite subset $L$ of $\mathbf{N}$ which does not meet $H_1$, we have $\|e - p_L\| \leq \varepsilon (\|p_{H_0}\|)^{-1}$. As above, it follows that, for each finite subset $H \supset H_1$,

Cauchy’s criterion therefore shows that $J \to p_J$ has a limit in $A$ with respect to the directed set $\mathfrak{F}(\mathbf{N})$.

If all the $x_n$ are units, then so are all the finite partial products $p_J$; hence for each finite subset $H$ containing $H_0$ we have

and this shows that, in the multiplicative group $G$ of units of $A$, the image under the mapping $J \to p_J$ of the section filter of $\mathfrak{F}(\mathbf{N})$ is a Cauchy filter base with respect to the left uniformity of $G$; but since $G$ is complete (Chapter IX, § 3, no. 7, Proposition 13), the limit of the mapping $J \to p_J$ belongs to $G$.