Remark

Proposition 13 is not necessarily true for a non-complete normed algebra. For example, in the algebra $C(I; \mathbf{R})$ of all finite continuous real-valued functions on $I = [0, 1]$ [the norm being $||x|| = \sup_{t \in I} |x(t)|$], the subalgebra $P$ of all polynomials in $t$ (restricted to $I$) is not complete; if $x(t)$ is any non-constant polynomial, then $1 + \varepsilon x$ is arbitrarily close to the identity element 1 of $P$ when $\varepsilon$ is arbitrarily small, but $1 + \varepsilon x$ is not a unit in $P$. However, if $A$ is a non-complete normed algebra, $G$ its group of units, $\hat{A}$ the completion of $A$, then $G$ is a subgroup of the group of units of $\hat{A}$, and therefore the topology induced on $G$ by the topology of $A$ is compatible with the group structure of $G$.