Remark

Let A be a commutative Banach algebra.

1) If $\chi$ is such that $\mathscr{G}'(x)$ vanishes in a neighbourhood of $\chi$, then $x$ belongs to I in a neighbourhood of $\chi$.

2) If $x$ belongs to I in a neighbourhood of $\chi$ and $y\in A$ is an arbitrary element, then $xy$ belongs to I in a neighbourhood of $\chi$.

3) The set of $\chi$ such that $x$ belongs to I in a neighbourhood of $\chi$ is open in $\mathsf{X}'(A)$.

4) Suppose that A is regular and without radical. If $x$ belongs to I in a neighbourhood of $\chi$ for every $\chi \in \mathsf{X}'(A)$, then $x$ belongs to I (Cor. 2 of I, p. 91 applied to the function $f=\mathscr{G}'(x)$ and Prop. 8 of I, p. 38).

5) Suppose that A is regular. Let I be an ideal of A and $\chi$ an element of $\mathsf{X}(A)$ such that $\chi \notin V(I)$. Then every element $x$ of A belongs to I in a neighbourhood of $\chi$. In fact, by Definition 1 of I, p. 89, there exists a $z\in A$ such that $\mathscr{G}'(z)$ is equal to 1 in a neighbourhood of $\chi$, and equal to 0 in a neighbourhood of V(I). The support of $\mathscr{G}(z)$ is compact and therefore one has $z\in I$ (Prop. 4 applied to V(I)), hence $xz\in I$, and $\mathscr{G}'(xz) =\mathscr{G}'(x)$ in a neighbourhood of $\chi$.

Recall that a subspace K of a topological space X is said to be perfect if it is closed and has no isolated point (TG, I, p. 8).