Show that a space X is homeomorphic to the space of multiplicative linear functionals Let $\mathcal{A}=C(X,\mathbb{R})$ where $X$ is a compact Hausdorff space. Let $\hat{\mathcal{A}}$ be equal to the set of multiplicative linear functionals from $\mathcal{A}$ to $\mathbb{R}$.
STATEMENT: Let $f\in\mathcal{A}$ and let $\hat{f}$ be the function from $\hat{\mathcal{A}}$ to $\mathbb{R}$ defined by $\hat{f}(\phi)=\phi(f)$. Put on $\hat{\mathcal{A}}$ the initial topology from the collection of all the functions $\hat{f}$ for $f\in\mathcal{A}$.  Prove that the map $x\mapsto \phi_x$ is a homeomorphism from $X$ onto $\hat{\mathcal{A}}$.
QUESTION: I have already shown that the given map is bijective and continuous, but I do not know how to show that the map is open. Any hints?
 A: I’m assuming that $\varphi_x$ is evaluation at $x$. Instead of just giving the argument or a hint, I’ll show how I actually worked my way to the argument.
Let $U$ be a non-empty open subset of $X$, and let $\Phi=\{\varphi_x:x\in U\}$; we want to show that $\Phi$ is open. You already know that the map is bijective, so any point of $\Phi$ is $\varphi_x$ for some $x\in U$. Fix $x\in U$; we want to find a finite $\{f_1,\ldots,f_n\}\subseteq\mathscr{A}$ and open sets $V_1,\ldots,V_n$ in $\Bbb R$ such that $\varphi_x\in\bigcap_{k=1}^n(\hat f_k)^{-1}[V_k]\subseteq\Phi$.
Now $\varphi_x\in(\hat f_k)^{-1}[V_k]$ iff $f_k(x)=\varphi_x(f_k)=\hat f_k(\varphi_x)\in V_k$, so we need only take $V_k$ to be an open nbhd of $f_k(x)$ for $k=1,\ldots,n$ in order to ensure that $\varphi_x\in\bigcap_{k=1}^n(\hat f_k)^{-1}[V_k]$.
For $y\in X$ we have $\varphi_y\in(\hat f_k)^{-1}[V_k]$ iff $f_k(y)\in V_k$, so $\bigcap_{k=1}^n(\hat f_k)^{-1}[V_k]\subseteq\Phi$ iff $\bigcap_{k=1}^nf_k^{-1}[V_k]\subseteq U$. Thus, it only remains to show that there are a finite $\{f_1,\ldots,f_n\}\subseteq\mathscr{A}$ and open nbhds $V_k$ of $f_k(x)$ for $k=1,\ldots,n$ such that $\bigcap_{k=1}^nf_k^{-1}[V_k]\subseteq U$.
In fact we require only one function: $X$, being compact Hausdorff, is Tikhonov, so there is an $f\in\mathscr{A}$ such that $f(x)=0$, and $f(x)=1$ for all $y\in X\setminus U$, and I expect that you can finish it off from here.
