Proposition 1

For every closed subset Φ of X, let $I_{\Phi}$ be the set of $f\in \mathscr{C}_0(X)$ that are zero on Φ. Then $\Phi \mapsto I_{\Phi}$ is a bijection of the set of closed subsets of X onto the set of closed ideals of $\mathscr{C}_0(X)$.

The set $I_{\Phi}$ is a closed ideal of $\mathscr{C}_0(X)$.

Let $\Phi \not= \Phi '$ be closed subsets of X. Exchanging Φ and $\Phi '$ if necessary, we may suppose that there exists $x\in \Phi '$ such that $x\notin \Phi$, and then there exists a function $f\in \mathscr{C}_0(X)$ which is zero on Φ and non-zero at $x$ (TG, IX, p. 43, prop. 1). We have $f\in I_{\Phi}$ and $f\notin I_{\Phi'}$, so that the mapping $\Phi \mapsto I_{\Phi}$ is injective.

Let I be a closed ideal of $\mathscr{C}_0(X)$. Let Φ be the set of $x\in X$ such that $f(x) = 0$ for every $f\in I$; this is a closed subset of X, and one has $I\subset I_{\Phi}$. Let us prove that $I_{\Phi}\subset I$, which will imply that $I = I_{\Phi}$ and complete the proof of the proposition.

Let $f\in I_{\Phi}$. For every real number $\varepsilon >0$, let $C_{\varepsilon}$ denote the set of $x\in X$ such that $|f(x)|\geqslant \varepsilon$. Since $f$ tends to 0 at infinity, the set $C_{\varepsilon}$ is compact. Let $x\in C_{\varepsilon}$; since $f(x)\not= 0$ and $f\in I_{\Phi}$, one has $x\notin \Phi$; by definition of Φ, there then exists a function $\varphi_x\in I$ such that $|\varphi_x(x)|>1$, hence such that $|\varphi_x(y)|>1$ for every $y$ belonging to a neighbourhood $V_x$ of $x$. The open sets $V_x\cap C_{\varepsilon}$ cover $C_{\varepsilon}$. Since the set $C_{\varepsilon}$ is compact, there exists a finite subset $T_{\varepsilon}\subset X$ such that

Then the element

1

$x\in T_{\varepsilon}$

of $\mathscr{C}_0(X)$ belongs to I, and one has $g_{\varepsilon}\geqslant \varepsilon^{-1}$ on $C_{\varepsilon}$. The function

belongs to I. For $x \notin C_{\varepsilon}$, one has

and for $x\in C_{\varepsilon}$, one has

Thus $f_{\varepsilon}$ converges uniformly to $f$ on X when $\varepsilon$ tends to 0. Hence $f\in \overline{I}$, whence $f\in I$ since I is closed.