4
$\begingroup$

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

  • 2
    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.2011-10-28
  • 2
    user16283 If @t.b.'s comment was sufficient, perhaps you can post (and accept) an answer yourself to check that you understood completely.2011-10-28

1 Answers 1