Gelfand-Naimark Theorem of non-unital case

Question

Let $X$ be a non-compact, locally compact Hausdorff space. Then $C_0 (X)$, the space of complex valued continuous functions on X vanishing at infinity, is a non-unital $C^*$-algebra and $\Omega (C_0 (X))$, the space of linear functionals $T$ on $C_0 (X)$ satisfying $T(fg)=T(f)T(g)$ for all $f,g \in C_0 (X)$, is a locally compact Hausdorff space with respect to weak${}^*$ topology in $C_0 (X)^*$. Let $x \in X$. Define $\hat{x}(f)=f(x)$ for all $f \in C_0 (X)$. Then $\hat{x} \in \Omega (C_0 (X))$. I know the map from $X$ to $\Omega (C_0 (X))$, $x \mapsto \hat{x}$, is bijective and continuous. How is the proof that the map is open map?

Answer 1

Let us write $F:X\to \Omega(C_0(X))$ for the map $F(x)=\hat{x}$. Let $U\subseteq X$ be open and $x\in U$. Let $V$ be open such that $x\in V\subseteq \overline{V}\subseteq U$ and $\overline{V}$ is compact. By Urysohn's lemma, there exists a continuous function $f:X\to [0,1]$ such that $f(x)=1$ and $f(y)=0$ for all $y\in X\setminus V$. Since $\overline{V}$ is compact, $f\in C_0(X)$. We now see that $\{\hat{y}\in\Omega(C_0(X)):\hat{y}(f)\neq 0\}$ is an open subset of $\Omega(C_0(X))$ which contains $\hat{x}$ and is contained in $F(U)$. Since $x\in U$ was arbitrary, this means $F(U)$ is open, and so $F$ is an open map.