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 $ell^inftyto mathbb{C}$. Note that, if $Bsubsetmathbb{N}$ and $chi_B$ is its characteristic function, then $varphi(chi_B)in {0,1}$ for any $varphiinPhi$ because

$$

varphi(chi_B)=varphi(chi_B^2)=(varphi(chi_B))^2

$$

Fix $varphiin Phi$, $varepsilon>0$, and $xin ell^infty$. Put

$$

A={ninmathbb{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 $nin mathbb{N}$ can be identified with the projection $xmapsto x_n$, 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 betamathbb N\

varphimapsto {Asubset mathbb N:varphi(chi_A)=1}

$$

and

$$

G:betamathbb{N}to Phi \

mathcal{U}mapsto left(xmapsto lim_mathcal{U} xright)

$$

The identity I’m trying to prove arises when checking that $(Gcirc F)(varphi)=varphi$ for every $varphiinPhi$. 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.