Let $M$ be the maximal ideal space of $L^{\infty} ([0,1],m)$ where $m$ is a Lebesgue measure.
I have shown that $L^{\infty} ([0,1],m)$ is a commutative $C^*$-algebra under essential supremum norm.
But how to show that the maximal ideal space has no isolated points?
Having difficulty in visualizing the maximal ideal space.
Ref: Rudin Functional Analysis Chapter 11 Problem 18.
If you already know Gelfand duality, the proof is easy enough.
Let $X$ be the maximal ideal space of $L^\infty([0,1])$. By Gelfand duality, $C(X)\cong L^\infty([0,1])$ as $C^\ast$-algebras. If $x\in X$ were an isolated point, then the characteristic function $1_x$ would be a minimal projection in $C(X)$, i.e., a projection that dominates no other non-zero projection. However, $L^\infty([0,1])$ does not contain any (non-zero) minimal projections.
As requested in the comments, let me elaborate a little on minimal projections. An element $p$ of a $C^\ast$-algebra $A$ is called projection if $p^\ast=p^2=p$. A projection $p$ is minimal if for every projection $q\in A$ such that $q\leq p$ either $q=0$ or $q=p$ holds (informally speaking, when it comes to non-zero projections, it doesn't get any smaller than $p$).
A function $f\in C(X)$ is a projection if $f$ takes values in $\{0,1\}$, that is, $f$ is a characteristic function. For a characteristic function to be continuous, both the preimages of $0$ and $1$ have to be open, that is, the function has to be the characteristic function of a clopen subset. A minimal projection in $C(X)$ is then a characteristic function of a clopen set that does not contain any other non-empty clopen set. Clearly this holds for a singleton containing an isolated point.
The projections in $L^\infty([0,1])$ are quite similar, you just have to remember that the elements of $L^\infty([0,1])$ are equivalence classes of functions rather than functions and the continuity condition is replaced by measurability. In short, the projections in $L^\infty([0,1])$ correspond to the measurable subsets of $[0,1]$ modulo null sets. There are no (non-zero) minimal projections in $L^\infty([0,1])$: Indeed, if $A\subset [0,1]$ is measurable with $m(A)>0$, then you can find a measurable set $B\subset A$ with $m(B)=\frac 1 2 m(A)$ (because $t\mapsto m([0,t]\cap A)$ is continuous) and get $0\lneq 1_B\lneq 1_A$.