Abelian C*-Algebras and the Functional Calculus in C*-Algebras

Prove a converse to the following proposition: if $\mathcal{A}$ is an abelian $C^*$-algebra with maximal ideal space $\Sigma$ and $a \in \mathcal{A}$ is such that $\mathcal{A} = C^*(a)$ (the $C^*$-algebra generated by $a$ and the identity), then the map $\tau : \Sigma \to \sigma (a)$ defined by $\tau (h) = h(a)$ is a homeomorphism. Specifically, show: if $K$ is a compact subset of $\mathbb {C}$, $C(K)$ is a singly generated $C^*$-algebra.

If $\mathcal{A}$ is an abelian $C^*$-algebra with a finite number of $C^*$-generators $a_1, \dots , a_n$, then there is a compact subset $X$ of $\mathbb {C}^n$ and an isometric $*$-isomorphism $\rho : \mathcal{A} \to C(X)$ such that $\rho (a_k) = z_k$, $1 \leq k \leq n$, where $z_k(\lambda_1, \dots , \lambda_n) = \lambda_k$ (see Exercise VII.8.12).

If $X$ is a compact Hausdorff space, show that $X$ is totally disconnected if and only if $C(X)$ is the closed linear span of its projections ($\equiv$ hermitian idempotents).

Using the terminology of the preceding exercise, show that if $(X, \Omega , \mu )$ is a $\sigma$-finite measure space, the maximal ideal space of $L^\infty (X, \Omega , \mu )$ is totally disconnected.

If $\mathcal{A}$ is a $C^*$-algebra with identity and $a = a^*$, show that $\exp (ia) = u$ is unitary. Is the converse true?

Let $X$ be compact and fix a point $x_0$ in $X$. Let

Show that $\mathcal{A}$ is an abelian $C^*$-algebra with identity and find its maximal ideal space.

If $X$ is completely regular, then $C_b(X)$ is a $C^*$-algebra and its maximal ideal space is the Stone--Cech compactification of $X$.