Proposition 2

Let A be a unital commutative regular Banach algebra. Let $n\geqslant 1$ be an integer and $(U_1, . . . ,U_n)$ an open covering of $\mathsf{X}(A)$. There exist elements $x_1, . . . , x_n$ of A of sum 1 such that Supp($\mathscr{G}(x_i)$)$\subset U_i$ for $i= 1, . . . ,n$.

Let us prove the proposition by induction on $n$. The assertion is valid if $n= 1$. Suppose that $n\geqslant 2$ and that the assertion is established for $n-1$.

There exists an open covering $(V_1, . . . ,V_n)$ of $\mathsf{X}(A)$ such that $\overline{V}_i\subset$ $U_i$ for all $i$. By the induction hypothesis, there exist elements $x, x_3, . . . , x_n\in A$ such that $x+x_3+\cdots +x_n= 1$ and Supp($\mathscr{G}(x)$)$\subset V_1\cup V_2$, Supp($\mathscr{G}(x_i)$)$\subset V_i$ for $i\geqslant 3$. Let K = Supp($\mathscr{G}(x)$)$\subset V_1\cup V_2$. Let $K_1$ (resp. $K_2$) be the set of elements of K which do not belong to $V_1$ (resp. $V_2$). Then $K_1$ and $K_2$ are disjoint compact subsets of K. Since the Banach algebra A is regular, there therefore exists $y\in A$ such that $\mathscr{G}(y) = 1$ on $K_1$ and $\mathscr{G}(y) = 0$ on $K_2$. Then $\mathscr{G}(xy)$ is zero on $\mathsf{X}(A)-K$ and on $K_2$, hence Supp $\mathscr{G}(xy)\subset \overline{V}_2\subset U_2$. Analogously, $\mathscr{G}(x(1-y))$ is zero on $\mathsf{X}(A)$ - K and on $K_1$, hence Supp $\mathscr{G}(x(1-y))\subset \overline{V}_1\subset U_1$. The elements $x_1=x(1-y),x_2=xy$, and $x_3, . . . ,x_n$ then satisfy the properties of the proposition.