Show that the maximal ideal space of $l^{\infty}$ is totally disconnected. I have this problem in my operator theory assignment:
Show that the maximal ideal space of $l^{\infty}$ is totally disconnected.
How to approach this . what does it mean to say maximal ideal space is totally disconnected.
What i know is that since $l^{\infty}$ is commutative unital Banach algebra its maximal ideal space is one to one correspondence  with its character space(set of all multiplicative linear functionals)
please provide some  hint.Thanks in Advanced
 A: Totally disconnected means that the connected components are the single points. 
You have that $\ell^\infty(\mathbb N)\simeq C(X)$, where $X$ is the maximal ideal space. It is well-known that $X=\beta\mathbb N$, the Stone-Cech compactification of $\mathbb N$, but we don't need that here. What we do need, is that $X$ is Hausdorff and compact, and thus normal (T4). 
Let $E\subset X$ be a connected component. Then $E$ is clopen and $1_E$ is continuous. By the above isomorphism, $1_E$ corresponds to some projection $p\in \ell^\infty(\mathbb N)$. This $p$ is necessarily minimal, since any decomposition $p=q+(p-q)$ would imply $1_E=1_F+1_G$ with $F,G$ closed and disjoint, so $E=F\cup G$ with $F,G$ clopen. The projections in $\ell^\infty(\mathbb N)$ are the sequences of $1$ and $0$, so to be minimal we must have $p=e_j$ for some $j$. 
If $E$ is not a single point, let $a,b\in E$ be distinct points. Because $X$ is T4, by Urysohn's Lemma there exists $f\in C(X)$ with $f(a)=1$, $f(b)=0$. Then $g=f\,1_E\in C(X)$. Let $y\in\ell^\infty(\mathbb N)$ be the element corresponding to $g$. As $0\leq g\leq 1_E$, we have $0\leq y\leq p=e_j$. But then $y=\alpha p$ for some $\alpha>0$, and thus $g=\alpha 1_E$. But now
$$
1=g(a)=\alpha 1_E(a)=\alpha1_E(b)=g(b)=0,
$$
a contradiction. 
It follows that $E$ consists of a single point. 

Edit: $X=\beta\mathbb N$.
The usual way to construct $\beta\mathbb N$ is as the weak$^*$-closure of $\delta(\mathbb N)$, where $\delta:\mathbb N\to C_b(\mathbb N)^*$ is $\delta_n(x)=x_n$. Since $\mathbb N$ is discrete, we have $C_b(\mathbb N)=\ell^\infty(\mathbb N)$. If $\gamma\in\beta\mathbb N$, then there exists a net $\{n_j\}_j\subset\mathbb N$ such that $\gamma=\lim_j\delta_{n_j}$; being a limit of characters in a unital C$^*$-algebra, $\gamma$ is also a character. Thus $\beta\mathbb N\subset X$. 
It remains to that that $X\setminus\beta\mathbb N=\varnothing$; note that this inclusion occurs in $\ell^\infty(\mathbb N)^*$. Let $h\in X\setminus\beta\mathbb N$. By Hahn-Banach (geometric version, with the weak$^*$-topology) there exists $x$ in the weak$^*$-dual of $\ell^\infty(\mathbb N)^*$ with $x(h)=1$ and $x|_{\beta\mathbb N}=0$. The weak$^*$ double dual of $\ell^\infty(\mathbb N)$ is itself (this is true for any locally convex space). So $x\in \ell^\infty(\mathbb N)$, and now we have $h(x)=1$ and $\delta_n(x)=0$ for all $n\in\mathbb N$. This is a contradiction, since $\delta_n=0$ for all $n$ means that $x=0$. So $h$ does not exist, and $\beta\mathbb N=X$. 
