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
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 $\beta\mathbb N=X$.