‎Theorem ‎2.1.15 of‎ ‎Murphy's ‎book

Question

The relevant theorem:

Let $\Omega$ be a compact Hausdorff space, and for each $\omega\in\Omega$ let $\delta_\omega$ be the character on $C(\Omega)$ given by evaluation at $\omega$; that is, $\delta_\omega(f)=f(\omega)$. Then the map $$\Omega\to\Omega(C(\Omega)),\qquad \omega\mapsto\delta_\omega,$$ is a homeomorphism.

When try show surjectivity, I don't understand how use Stone-Weierstrass to show that there exists a point $\omega\in\Omega$ such that $f(\omega)=0$ for all $f\in M$.

Because $M=\ker\tau\neq C(\Omega)$, so use neccesary $M$ don't have any constant?

thanks

Answer 1

I think you're not looking at the same version of the Stone-Weierstrass theorem that Murphy is. This is the relevant version:

If $\Omega$ is a locally compact Hausdorff space, and $M\subset C_0(\Omega)$ is a closed $*$-subalgebra, then $M=C_0(\Omega)$ if $M$ separates points and vanishes nowhere.

Since we know $M$ is a proper $*$-subalgebra and separates points, it must vanish somewhere.

Answer 2

It is one of the Stone-Weierstrass theorems, see corollary V.8.2, a course in functional analysis by Conway.

An alternative proof without using the Stone-Weierstrass theorem can be found here.