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

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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
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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

1 Answer 1

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.

Credits goes to t.b.

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