Characters on $\ell^\infty$ and characteristic functions Let $\ell^\infty$ be the set of bounded complex-valued sequences, which we regard as a commutative $C^*$ algebra under pointwise multiplication. Write $\Phi$ for the set of characters of $\ell^\infty$, i.e. non-zero algebra homomorphisms $\ell^\infty\to \mathbb{C}$. Note that, if $B\subset\mathbb{N}$ and $\chi_B$ is its characteristic function, then $\varphi(\chi_B)\in \{0,1\}$ for any $\varphi\in\Phi$ because
$$
\varphi(\chi_B)=\varphi(\chi_B^2)=(\varphi(\chi_B))^2
$$
Fix $\varphi\in \Phi$, $\varepsilon>0$, and $x\in \ell^\infty$. Put
$$
A=\{n\in\mathbb{N}:|x_n-\varphi(x)|<\varepsilon\}
$$
I'd like to show that $\varphi(\chi_A)=1$.
Some context: By the Gelfand-Naimark Theorem, $C(\Phi)\cong \ell^\infty$, and $n\in \mathbb{N}$ can be identified with the projection $x\mapsto x_n$, which obviously belongs to $\Phi$. Under this identification, $\mathbb N$ is dense in $\Phi$ and so $\Phi$ can be regarded as the Stone-Cech compacfication of $\mathbb N$. If we let $\beta \mathbb N$ be the set of ultrafilters on $\mathbb N$, then we can find explicit bijections between $\Phi$ and $\beta \mathbb N$ by considering
$$
F: \Phi \to \beta\mathbb N\\
\varphi\mapsto \{A\subset \mathbb N:\varphi(\chi_A)=1\}
$$
and
$$
G:\beta\mathbb{N}\to \Phi \\
\mathcal{U}\mapsto \left(x\mapsto \lim_\mathcal{U} x\right)
$$
The identity I'm trying to prove arises when checking that $(G\circ F)(\varphi)=\varphi$ for every $\varphi\in\Phi$.  Of course, one can also verify that they are continuous and then check that the compositions are the identity on $\mathbb{N}$, which as a by product yields what we want. However, I suspect there's a more direct, elementary argument that I'm not seeing.
 A: I'm not sure if this is what you're looking for, but I tried.
The evaluation characters $x\mapsto x(n)$ are of course $w^*$-dense in the set $\beta \mathbb{N} \subset (\ell^\infty)^*$ of all characters.
Therefore, given any $\varphi\in\beta\mathbb{N}$ and $x,y \in \ell^\infty$, there exists a sequence $n_k$ such that $\varphi(x) = \lim_k x(n_k)$ and $\varphi(y) = \lim_k y(n_k)$.
In particular, given $\varphi$ and $x$, suppose $\varphi(x) = c$, and let $y$ be the characteristic function of  $A = \{n : |x(n) - c| < \varepsilon\}$.
Since $c = \lim_{k} x(n_k)$, eventually the sequence $n_k$ lies inside $A$.
Therefore $\varphi(y) = \lim_k y(n_k) = 1$.
A: Let $$y_n=\begin{cases} \frac1{x_n-\varphi(x)}\quad n\notin A,\\0 \quad\,\qquad n\in A.\end{cases}$$
Then $|y_n|\leq \frac1\epsilon$ for all $n$ so $y\in \ell^\infty$.
Then
\begin{eqnarray*}1-\varphi(\chi_A)&=&\varphi(1-\chi_A)\\&=&\varphi((x-\varphi(x))y)\\&=&\varphi(x-\varphi(x))\varphi(y)\\&=&(\varphi(x)-\varphi(x))\varphi(y)\\&=&0.\end{eqnarray*}
