The following is a theorem about self-adjoint subalgebra of $C(X)$ where $X$ is compact Hausdorff and the first half of its proof:
Here are my questions:
- Why $\Gamma:\mathfrak{U}\to C(M_{\mathfrak U})$ being an isometry implies that $\Gamma(\mathfrak{U})$ is closed in $C(M_{\mathfrak U})$?
- Why does $\Gamma(\mathfrak{U})$ separate points?
Here $M_\mathfrak{U}$ denotes the set of non zero complex multiplicative linear functions on $\mathfrak{U}$. For the second question, I think the book means $\Gamma(\mathfrak{U})$ separate points of $M_{\mathfrak U}$, which means for any two distinct points $x,y\in M_{\mathfrak U}$, $f(x)\neq f(y)$ for some $f\in \Gamma(\mathfrak{U})$. But I don't see how to give an argument.
As Gro-Tsen commented, the fact that $\Gamma$ is an isometry implies its image is closed because $\mathfrak{U}$ is a complete metric space (since it is closed in $C(X)$). Any subset of a metric space that is complete as a metric space is automatically closed, since a sequence that converged to a point outside the set would be a Cauchy sequence with no limit in the set.
The fact that $\Gamma(\mathfrak{U})$ separates points comes from the fact that elements of $M_\mathfrak{U}$ are defined as functions on $\mathfrak{U}$. That is, if $x,y\in M_\mathfrak{U}$ are distinct, there must be some $u\in \mathfrak{U}$ such that $x(u)\neq y(u)$. But $x(u)=(\Gamma(u))(x)$ and similarly for $y$, so this just says that $\Gamma(u)$ separates $x$ from $y$.