a condition for a $C^*$ algebra to have a unit?

Question

I am looking for a hint or a methodology to approach this problem, showing in Arveson's A short course on spectral theory:

Let $X$ be a compact Hausdorff space and let $F$ be a proper closed subset of $X$. Let $A$ be the ideal of all functions $f\in C(X)$ that vanish throughout $F$ ($f(p)= 0\ \forall p ∈ F$). Note that $A$ is a $C^∗$-algebra in its own right.

a/ Show that $A$ has a unit if and only if $F$ is both closed and open.

$A$ is unital is equivalent to $\exists e\in A: fe=ef= f, \ \forall f\in A$
then we need $e$ to be satisfying $e(x)= 0\ \forall x\in F$ but $e(x)= 1 \ \forall x\in F^c$ then this implies that $F$ should be empty.
Is that not correct? if otherwise $F$ is not empty then there should be some $x\in F^c$ for which $e(x)= 1/2$ or how would a continuous function jump from $0$ to $1$, by Urysohn lemma, in fact $X$ is compact Hausdorff hence a normal space.
Actually if $X$ is not a discrete space, here a finite set, then the only open and closed subsets of $X$ are $\emptyset$ and $X$ itself, no ?

b/ Assuming that $F$ is not open, identify the unitalization of $A$ in concrete terms by exhibiting a compact Hausdorff space $Y$ such that $\tilde{A}\cong C(Y)$

How would I even approach this ? I assume that $A\cong C(sp(A))$ and that $sp(A)=\{\delta_x|\ x\in F^c\}$ the evaluation functionals

Answer 1

I'll discuss the answer to a). It should give some hints about b).

a) If $F$ is clopen (i.e. open and closed), then $e = \chi_F$ (i.e. the indicator function on $F$) is a unit for $A$. Clearly $e$ vanishes on $F$ and would be a unit for $A$ if it was in $A$, so all that we need to do is show that $e$ is continuous. If $U$ is any open subset of $\mathbb{C}$ (or indeed any subset of $\mathbb{C}$ at all), then $e^{-1} U$ is one of $\emptyset, F, F^\complement, X$, and if $F$ is clopen, then all of those sets are open. Thus if $F$ is clopen, then $A$ is unital.

Conversely, suppose that $A$ is unital, and $e \in A$ is the unit. Then $e \vert_F \equiv 0$, since $e \in A$. Now let $x \in F^\complement$. Then there exists a function $f \in A$ such that $f(x) \neq 0$. Therefore $e(x) = 1$ in order to have $e(x) f(x) = f(x)$. This means that $F^\complement = \{ x : e(x) = 1 \}$ must be closed, since $e$ is continuous. Thus if $A$ is unital, then $F$ is clopen.