Ytukyg

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?

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