# Existence of faithful state in $C^\ast$-algebras

Why does there always exist a faithful state in a separable $C^\ast$-algebra?

-
Well, since the unit ball in the dual space is weak$^{\ast}$-separable, there is a separating sequence of states $\omega_n$, then just sum them up $\omega = \sum 2^{-n} \omega_n$ and check that you get a faithful state. –  t.b. Oct 28 '11 at 19:41
user16283 If @t.b.'s comment was sufficient, perhaps you can post (and accept) an answer yourself to check that you understood completely. –  Srivatsan Oct 28 '11 at 20:58

Well, since the unit ball in the dual space is weak-$^*$-separable, there is a separating sequence of states $\omega_n$, then just sum them up $\omega=\sum 2^{-n}\omega_n$ and check that you get a faithful state.